🧠:[<think>Okay, so I need to solve this problem where we have a quartic polynomial f(x) with some given values. Let me try to recall what a quartic polynomial is. It's a polynomial of degree 4, right? So it can have up to four roots, but here they mention that f(1)=f(2)=f(3)=0, which suggests that 1, 2, and 3 are roots of the polynomial. But since it's a quartic, there should be one more root, or maybe a repeated root. Hmm.But wait, the problem doesn't state that these are the only roots. So maybe the polynomial can be written as f(x) = (x-1)(x-2)(x-3)(ax + b), where ax + b is a linear factor because we need a quartic. Or maybe there is a squared term? Let me check. If there are three distinct roots, 1, 2, 3, each with multiplicity 1, then the polynomial would have (x-1)(x-2)(x-3) as a factor, and then another linear factor. But since the degree is 4, that would make sense. So f(x) = (x-1)(x-2)(x-3)(cx + d), where c and d are constants to be determined.But we also have f(4) = 6 and f(5) = 72. So maybe we can use these two conditions to solve for c and d. Let me write that down.First, express f(x) as (x-1)(x-2)(x-3)(cx + d). Then plug in x=4:f(4) = (4-1)(4-2)(4-3)(4c + d) = 3*2*1*(4c + d) = 6*(4c + d) = 6.So 6*(4c + d) = 6 implies that 4c + d = 1. (Equation 1)Similarly, plug in x=5:f(5) = (5-1)(5-2)(5-3)(5c + d) = 4*3*2*(5c + d) = 24*(5c + d) = 72.Therefore, 24*(5c + d) = 72 implies that 5c + d = 3. (Equation 2)Now we have two equations:1) 4c + d = 12) 5c + d = 3Subtract equation 1 from equation 2:(5c + d) - (4c + d) = 3 - 1This simplifies to c = 2.Then substitute c = 2 into equation 1:4*2 + d = 1 => 8 + d = 1 => d = -7.Therefore, the polynomial is f(x) = (x-1)(x-2)(x-3)(2x -7).But wait, let me check if that works. Let's compute f(4):(4-1)(4-2)(4-3)(2*4 -7) = 3*2*1*(8 -7) = 6*1 = 6. Correct.f(5): (5-1)(5-2)(5-3)(2*5 -7) = 4*3*2*(10 -7) = 24*3 = 72. Correct. Okay, so that seems right.But let me just verify that this is indeed a quartic. The factors (x-1)(x-2)(x-3)(2x -7) multiply out to a degree 4 polynomial. So that's correct.But now the problem is asking for the last digit of f(2010). So I need to compute f(2010) modulo 10, which is the same as finding the last digit of f(2010).But computing f(2010) directly would be tedious because 2010 is a large number. However, since we only need the last digit, we can compute f(2010) modulo 10. Let me recall that modulo 10 arithmetic can simplify calculations.First, note that 2010 modulo 10 is 0. So x = 2010 is equivalent to x ≡ 0 mod 10.Therefore, let's compute f(0) modulo 10? Wait, no. Wait, f(2010) modulo 10. Let me think.But f(x) = (x -1)(x -2)(x -3)(2x -7). So f(2010) mod 10 is equivalent to computing each factor mod 10, multiplying them together, and then taking the result mod 10.But since 2010 ≡ 0 mod 10, let's compute each term:x = 2010 ≡ 0 mod 10Therefore:(x -1) ≡ 0 -1 ≡ -1 ≡ 9 mod 10(x -2) ≡ 0 -2 ≡ -2 ≡ 8 mod 10(x -3) ≡ 0 -3 ≡ -3 ≡ 7 mod 10(2x -7) ≡ 2*0 -7 ≡ -7 ≡ 3 mod 10Therefore, the four factors modulo 10 are 9, 8, 7, 3. Multiplying these together:First, 9 * 8 = 72 ≡ 2 mod 10Then, 2 * 7 = 14 ≡ 4 mod 10Then, 4 * 3 = 12 ≡ 2 mod 10Therefore, f(2010) ≡ 2 mod 10, so the last digit is 2. Wait, is that correct? Let me double-check each step.First, x=2010. Let's compute each term:x -1 = 2010 -1 = 2009. 2009 mod 10 is 9.x -2 = 2010 -2 = 2008. 2008 mod 10 is 8.x -3 = 2010 -3 = 2007. 2007 mod 10 is 7.2x -7 = 2*2010 -7 = 4020 -7 = 4013. 4013 mod 10 is 3.Therefore, the four terms mod 10 are 9, 8, 7, 3.Multiply them step by step:9 * 8 = 72. 72 mod 10 is 2.2 * 7 = 14. 14 mod 10 is 4.4 * 3 = 12. 12 mod 10 is 2.Yes, so indeed, the product is 2 mod 10. Therefore, the last digit of f(2010) is 2.Wait a second, but let me check if the polynomial is correctly defined. The polynomial is f(x) = (x-1)(x-2)(x-3)(2x -7). So when x=2010, this is equal to (2009)(2008)(2007)(2*2010 -7). But 2*2010 is 4020, minus 7 is 4013. So the product is 2009*2008*2007*4013. But computing that directly would give a huge number, but we only need the last digit. So by using modular arithmetic, as above, we can compute each term mod 10, then multiply them mod 10. Which gives 2.But let me verify once again. Suppose I compute (9 * 8 * 7 * 3) mod 10. 9*8=72 mod10=2, 2*7=14 mod10=4, 4*3=12 mod10=2. Yep. So that's 2.But let me check if there's another way this could be approached. Since the polynomial is quartic, maybe we can write f(x) as (x-1)(x-2)(x-3)(ax + b) and found a=2, b=-7. Alternatively, is there a chance that there's a repeated root? But the problem didn't specify multiplicity, just f(1)=f(2)=f(3)=0. So they are roots, but maybe not necessarily all simple roots. However, since the polynomial is quartic, if all three roots 1,2,3 are simple, then the remaining factor is linear. If there was a multiple root, then the remaining factor might be quadratic. But in this problem, they found a linear factor, so that suggests that 1,2,3 are all simple roots, and the fourth root is 7/2. So that seems okay.Alternatively, maybe the problem is expecting another approach. For example, since the polynomial is quartic and we have three roots, and two other points, we can set up equations to find the coefficients. But that would involve more variables. However, the approach above seems straightforward and gives a consistent answer. Let me see.Alternatively, maybe the polynomial can be written as (x-1)(x-2)(x-3)(px + q) + something. Wait, but in this problem, since f(1)=f(2)=f(3)=0, the polynomial must be divisible by (x-1)(x-2)(x-3), so it's (x-1)(x-2)(x-3)*quadratic, but no, it's quartic, so it's (x-1)(x-2)(x-3)*(linear). So the approach we took is correct.So I think the answer is 2. But let me check with another method just to be sure.Another approach: Since we need f(2010) mod 10, and 2010 ≡ 0 mod 10, we can compute f(0) mod 10. Wait, is that the case? Wait, f(2010) mod10 is equivalent to f(0) mod10, since 2010 ≡0 mod10.Wait, but f(x) is (x-1)(x-2)(x-3)(2x -7). So f(0) would be (-1)(-2)(-3)(-7) = (-1*-2)=2, (2*-3)= -6, (-6*-7)=42. So 42. 42 mod10 is 2. Therefore, f(0) mod10 is 2, so f(2010) mod10 is 2. Therefore, the last digit is 2.Wait, that's a simpler way to see it. Since f(2010) ≡ f(0) mod10, because 2010 ≡0 mod10. So compute f(0) mod10. So f(0) = (-1)(-2)(-3)(-7) = (-1*-2)=2, 2*(-3)= -6, -6*(-7)=42. 42 mod10 is 2. So same answer.Therefore, the last digit is 2. That seems consistent. So I think the answer is 2.But let me make sure there are no errors in my reasoning. Let's confirm once again.First, confirm the polynomial:We have f(x) = (x-1)(x-2)(x-3)(2x -7). Let's compute f(0):(-1)(-2)(-3)(-7) = (-1*-2) = 2; (2*-3) = -6; (-6*-7) = 42. So f(0)=42. 42 mod10=2. So f(2010)≡f(0)≡2 mod10. Therefore, the last digit is 2. That's correct.Alternatively, if we didn't use this modular approach, and tried to compute f(2010), it would be a huge number, but the last digit is determined by the operations modulo 10, so regardless of the size, the last digit is 2.Hence, the answer is 2. Therefore, the last digit of f(2010) is 2.Final Answerboxed{2}</think>Given a quartic polynomial ( f(x) ) with roots at ( x = 1, 2, 3 ), we can express it as ( f(x) = (x-1)(x-2)(x-3)(ax + b) ). We need to determine the coefficients ( a ) and ( b ) using the given values ( f(4) = 6 ) and ( f(5) = 72 ).First, substituting ( x = 4 ):[f(4) = (4-1)(4-2)(4-3)(4a + b) = 3 cdot 2 cdot 1 cdot (4a + b) = 6(4a + b) = 6]This simplifies to:[4a + b = 1 quad text{(Equation 1)}]Next, substituting ( x = 5 ):[f(5) = (5-1)(5-2)(5-3)(5a + b) = 4 cdot 3 cdot 2 cdot (5a + b) = 24(5a + b) = 72]This simplifies to:[5a + b = 3 quad text{(Equation 2)}]Subtracting Equation 1 from Equation 2:[(5a + b) - (4a + b) = 3 - 1 implies a = 2]Substituting ( a = 2 ) into Equation 1:[4(2) + b = 1 implies 8 + b = 1 implies b = -7]Thus, the polynomial is ( f(x) = (x-1)(x-2)(x-3)(2x - 7) ).To find the last digit of ( f(2010) ), we compute ( f(2010) mod 10 ). Since ( 2010 equiv 0 mod 10 ), we evaluate the polynomial at ( x = 0 mod 10 ):[f(0) = (-1)(-2)(-3)(-7) = (-1 cdot -2) cdot (-3 cdot -7) = 2 cdot 21 = 42]The last digit of 42 is 2. Therefore, the last digit of ( f(2010) ) is:[boxed{2}],<think>Okay, let's see. The problem is about a quartic polynomial ( f(x) ), which means it's a fourth-degree polynomial. They tell us that ( f(1) = f(2) = f(3) = 0 ), so those are roots of the polynomial. Then they give us ( f(4) = 6 ) and ( f(5) = 72 ). The question is asking for the last digit of ( f(2010) ).First, since it's a quartic polynomial (degree 4) and we have three roots at x=1, x=2, and x=3, the polynomial should have factors of ( (x-1)(x-2)(x-3) ). But since it's a quartic, there must be another linear factor, but since we only have three roots given, maybe there's a repeated root or another factor that's quadratic. Wait, but the problem doesn't say anything about other roots, so maybe it's written as ( f(x) = (x-1)(x-2)(x-3)(ax + b) ), where ( ax + b ) is a linear term, making it quartic. That seems possible. Then, we can use the given points ( f(4) = 6 ) and ( f(5) = 72 ) to solve for coefficients ( a ) and ( b ).Alternatively, maybe it's written as ( f(x) = (x-1)(x-2)(x-3)(cx + d) ). Let's check that. If we set up equations for f(4) and f(5):At x=4: ( f(4) = (4-1)(4-2)(4-3)(4c + d) = 3*2*1*(4c + d) = 6*(4c + d) = 6 ). So 6*(4c + d) =6. Dividing both sides by 6: 4c + d =1.Similarly, at x=5: ( f(5) = (5-1)(5-2)(5-3)(5c + d) =4*3*2*(5c + d) =24*(5c + d) =72. So 24*(5c + d)=72. Dividing both sides by 24: 5c + d=3.Now, we have a system of equations:1. 4c + d =12. 5c + d =3Subtract the first equation from the second: (5c + d) - (4c + d) =3 -1 => c=2. Then substitute back into first equation: 4*2 +d=1 =>8 +d=1 =>d= -7.So, the polynomial is ( f(x)= (x-1)(x-2)(x-3)(2x -7) ). Let me verify this with f(4) and f(5):At x=4: (3)(2)(1)(8 -7)=6*1=6. Correct.At x=5: (4)(3)(2)(10 -7)=24*3=72. Correct.Okay, so that works. Therefore, the polynomial is determined. Now, we need to find the last digit of f(2010). Since we need only the last digit, we can compute f(2010) modulo 10. That would give the last digit.So, instead of computing f(2010) directly, which might be a huge number, we can compute each term modulo 10 and then multiply them together modulo 10.Let's write f(2010) = (2010 -1)(2010 -2)(2010 -3)(2*2010 -7) = (2009)(2008)(2007)(4020 -7) = (2009)(2008)(2007)(4013).But since we need modulo 10, let's compute each factor modulo 10:First term: 2010 -1 =2009. 2009 mod10=9.Second term:2010 -2=2008. 2008 mod10=8.Third term:2010 -3=2007. 2007 mod10=7.Fourth term:2*2010 -7=4020 -7=4013. 4013 mod10=3.Therefore, modulo 10, f(2010) ≡9*8*7*3 mod10.Compute step by step:9*8=72, 72 mod10=2.2*7=14, 14 mod10=4.4*3=12, 12 mod10=2.Therefore, the last digit is 2. Wait, but let me check again. Wait, but maybe the polynomial is written as f(x) = (x-1)(x-2)(x-3)(2x -7). So, when x=2010, each term is (2010 -1), (2010 -2), (2010 -3), and (2*2010 -7). So, yes, 2009,2008,2007,4013. Each mod10 gives 9,8,7,3. Multiply them: 9*8=72 mod10=2, 2*7=14 mod10=4, 4*3=12 mod10=2. So, final answer is 2. Therefore, the last digit is 2. Hmm. But wait, let me check again.Alternatively, maybe the mistake is here. Wait, when we write the polynomial as (x-1)(x-2)(x-3)(2x -7), when x=2010, that's (2010 -1)(2010 -2)(2010 -3)(2*2010 -7). So, each term is 2009,2008,2007,4013. Modulo10, each is 9,8,7,3. Multiply them all together mod10:9 *8=72; 72 mod10=2Then 2 *7=14; 14 mod10=4Then 4 *3=12; 12 mod10=2So, yes, 2. Therefore, the last digit is 2.But let me check once again. Let me compute f(2010) mod10 step by step.But maybe there is another way. Let's think again. The polynomial is (x-1)(x-2)(x-3)(2x -7). So, perhaps expand the polynomial and then compute each coefficient mod10, then evaluate at x=2010 mod10=0. Wait, 2010 mod10=0, so x=0 mod10.But maybe that's a different approach. Let's try that. Let me expand the polynomial f(x) modulo10. Let's compute each coefficient mod10.First, compute (x-1)(x-2)(x-3):First, multiply (x-1)(x-2)=x^2 -3x +2.Then multiply by (x-3):(x^2 -3x +2)(x -3)=x^3 -3x^2 -3x^2 +9x +2x -6= x^3 -6x^2 +11x -6.So, (x-1)(x-2)(x-3)=x^3 -6x^2 +11x -6.Now, multiply by (2x -7):Multiply each term:x^3*(2x -7)=2x^4 -7x^3-6x^2*(2x -7)= -12x^3 +42x^211x*(2x -7)=22x^2 -77x-6*(2x -7)= -12x +42Combine all terms:2x^4 -7x^3 -12x^3 +42x^2 +22x^2 -77x -12x +42Combine like terms:2x^4 + (-7 -12)x^3 + (42 +22)x^2 + (-77 -12)x +42=2x^4 -19x^3 +64x^2 -89x +42Now, reduce each coefficient mod10:2x^4 -19x^3 +64x^2 -89x +42mod10:2x^4 + (-19 mod10)x^3 + (64 mod10)x^2 + (-89 mod10)x + (42 mod10)Compute each:-19 mod10: -19 +20=164 mod10=4-89 mod10: -89 +90=142 mod10=2So, polynomial mod10 is:2x^4 +1x^3 +4x^2 +1x +2So, f(x) ≡2x^4 +x^3 +4x^2 +x +2 mod10.Now, evaluate at x=2010. Since 2010 mod10=0, so x=0 mod10.Therefore, f(0) mod10=2*(0)^4 +0^3 +4*(0)^2 +0 +2=0 +0 +0 +0 +2=2 mod10.Therefore, the last digit is 2. So, either way, the answer is 2.But wait, the previous method also gave 2, so that's consistent. Therefore, the last digit is 2.But just to make sure, maybe there is a mistake in the coefficients when expanding. Let me check the expansion again.First, (x-1)(x-2)(x-3):First, (x-1)(x-2)=x² -3x +2.Then (x² -3x +2)(x-3)=x³ -3x² -3x² +9x +2x -6= x³ -6x² +11x -6. Correct.Multiply by (2x -7):x³*(2x -7)=2x⁴ -7x³-6x²*(2x -7)= -12x³ +42x²11x*(2x -7)=22x² -77x-6*(2x -7)= -12x +42Combine:2x⁴ -7x³ -12x³ +42x² +22x² -77x -12x +42=2x⁴ -19x³ +64x² -89x +42. Correct.Modulo10: 2x⁴ + (-19 mod10)x³ +64 mod10 x² + (-89 mod10)x +42 mod10=2x⁴ +1x³ +4x² +1x +2. Correct.Then f(0)=2*0 +0 +0 +0 +2=2. So yes. Therefore, the answer is 2. That seems correct.Alternatively, if you use the original factors:(2009)(2008)(2007)(4013) mod10Each term mod10: 9,8,7,3.Multiply them: 9*8=72 mod10=2; 2*7=14 mod10=4; 4*3=12 mod10=2.Same result.Therefore, the last digit is 2. So the answer is 2. So, in the box, boxed{2}Final Answerboxed{2}</think>Given a quartic polynomial ( f(x) ) with roots at ( x = 1 ), ( x = 2 ), and ( x = 3 ), we can express ( f(x) ) as ( (x-1)(x-2)(x-3)(ax + b) ). Using the given values ( f(4) = 6 ) and ( f(5) = 72 ), we solve for the coefficients ( a ) and ( b ).1. For ( x = 4 ): [ f(4) = (4-1)(4-2)(4-3)(4a + b) = 3 cdot 2 cdot 1 cdot (4a + b) = 6(4a + b) = 6 ] Simplifying, we get ( 4a + b = 1 ).2. For ( x = 5 ): [ f(5) = (5-1)(5-2)(5-3)(5a + b) = 4 cdot 3 cdot 2 cdot (5a + b) = 24(5a + b) = 72 ] Simplifying, we get ( 5a + b = 3 ).Solving the system of equations:- Subtracting the first equation from the second: ( (5a + b) - (4a + b) = 3 - 1 ) gives ( a = 2 ).- Substituting ( a = 2 ) into ( 4a + b = 1 ): ( 8 + b = 1 ) gives ( b = -7 ).Thus, the polynomial is ( f(x) = (x-1)(x-2)(x-3)(2x - 7) ).To find the last digit of ( f(2010) ), we compute ( f(2010) mod 10 ):- Each factor modulo 10: - ( 2010 - 1 equiv 9 mod 10 ) - ( 2010 - 2 equiv 8 mod 10 ) - ( 2010 - 3 equiv 7 mod 10 ) - ( 2 cdot 2010 - 7 equiv 3 mod 10 )Multiplying these results modulo 10:[9 cdot 8 cdot 7 cdot 3 mod 10 = (72 mod 10) cdot 7 cdot 3 = 2 cdot 7 cdot 3 mod 10 = (14 mod 10) cdot 3 = 4 cdot 3 mod 10 = 12 mod 10 = 2]Thus, the last digit of ( f(2010) ) is (boxed{2}).]

🔑:【Analysis】It is given in the problem that 1,2,3 are three zeros of the function f(x), so we consider starting from the zero point form of the function.Solution From the problem, we knowf(1)=f(2)=f(3)=0 text {. }Let the one-variable quartic polynomial function bef(x)=(x-1)(x-2)(x-3)(a x+b) text {, }where a, b in mathbf{R}.Therefore, left{begin{array}{l}f(4)=6(4 a+b)=6, f(5)=24(5 a+b)=72 .end{array}right.Solving these, we get a=2, b=-7.Thus, f(x)=(x-1)(x-2)(x-3)(2 x-7).Hence, f(2010)begin{array}{l}=(2010-1)(2010-2)(2010-3)(4020-7) equiv 2(bmod 10) .end{array}Therefore, the last digit of f(2010) is 2.

🧠:[<think>Okay, so I have this problem here: given positive real numbers x, y, z that satisfy the equation xyz(x + y + z) = 1, I need to find the minimum value of (x + y)(x + z). Hmm, alright. Let me think about how to approach this. First off, I remember that optimization problems like this often involve using inequalities such as AM-GM or Cauchy-Schwarz. Maybe I can apply those here? Let me recall the AM-GM inequality: for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. That might come in handy since all variables here are positive. The expression we need to minimize is (x + y)(x + z). Let me expand that to see if it simplifies. Multiplying it out, we get:(x + y)(x + z) = x² + xz + xy + yz.Hmm, so that's x squared plus x times z plus x times y plus y times z. Maybe I can rewrite this expression in terms of the variables given and the constraint. The constraint is xyz(x + y + z) = 1. That seems a bit complicated, but perhaps there's a substitution or a way to relate the two.Let me denote S = x + y + z. Then the constraint becomes xyzS = 1. So S = 1/(xyz). Hmm, maybe I can express S in terms of xyz here, but I'm not sure yet. Let's see. Alternatively, maybe I can use substitution. Let's see if setting variables to be equal might help. Sometimes symmetry in variables can lead to extrema. The problem is that x, y, z are all variables here, but the expression (x + y)(x + z) isn't symmetric in all variables—it treats x differently from y and z. So perhaps the minimum occurs when y and z are equal? That might be a common scenario in optimization problems with partial symmetry.Let me try assuming that y = z. Maybe that simplifies things. If I set y = z, then the constraint becomes x * y² * (x + 2y) = 1. The expression to minimize becomes (x + y)(x + y) = (x + y)². Wait, is that right? If y = z, then (x + y)(x + z) = (x + y)(x + y) = (x + y)². So we need to minimize (x + y)² given that x * y² * (x + 2y) = 1. That seems manageable. Let's try to work with that.Let me denote t = x + y. Then, x = t - y. Substituting into the constraint: (t - y) * y² * (t - y + 2y) = (t - y) * y² * (t + y) = (t² - y²) * y² = 1. So we have (t² - y²)y² = 1. But we need to minimize t². Hmm, this seems a bit tangled. Maybe another substitution. Let me set u = y². Then, (t² - u)u = 1. So, t² = (1/u) + u. Therefore, t² = u + 1/u. The minimum of u + 1/u is 2, achieved when u = 1. So if u = 1, then t² = 2, so t = sqrt(2). Therefore, the minimal value of (x + y)² would be 2, so (x + y)(x + z) = 2. Wait, but is this valid?Wait, hold on. Let's check the substitution steps again. If u = y², then (t² - u)u = 1. Then, t² = (1/u) + u. So, t² is indeed u + 1/u. Then, the minimal value of t² is 2 when u = 1. So that would mean y² = 1, so y = 1 (since y is positive). Then, substituting back into the constraint, when y = 1, (t² - 1)(1) = 1, so t² - 1 = 1, so t² = 2, so t = sqrt(2). Then x + y = sqrt(2), so x = sqrt(2) - y = sqrt(2) - 1. Then, check if this satisfies the original constraint.Original constraint: xyz(x + y + z). If y = z = 1, x = sqrt(2) - 1. Then x + y + z = (sqrt(2) - 1) + 1 + 1 = sqrt(2) + 1. Then, xyz = (sqrt(2) - 1)*1*1 = sqrt(2) - 1. Then, xyz(x + y + z) = (sqrt(2) - 1)(sqrt(2) + 1) = (sqrt(2))² - 1² = 2 - 1 = 1. Perfect! So that satisfies the constraint. So in this case, (x + y)(x + z) = (sqrt(2) - 1 + 1)^2 = (sqrt(2))² = 2. So the minimum value is 2? But wait, the question is to find the minimum of (x + y)(x + z). If we assume y = z, we get 2. But is this the actual minimum, or could there be a lower value when y ≠ z?Hmm, good point. Maybe we need to check if the minimal value is indeed 2, even when y and z are not equal. Let me think. How can we approach this without assuming y = z?Let's consider the expression (x + y)(x + z). Let me write this as x² + x(y + z) + yz. Now, perhaps express this in terms of x and S = y + z, and yz. Let me denote S = y + z, P = yz. Then, the expression becomes x² + xS + P. The constraint is xyz(x + y + z) = xP(x + S) = 1. So, we need to minimize x² + xS + P given that xP(x + S) = 1.Hmm, variables here are x, S, P. But S and P are related since for positive y and z, S² ≥ 4P by AM-GM (since (y + z)/2 ≥ sqrt(yz), so S² ≥ 4P). So S² ≥ 4P. That might be useful.So perhaps we can write P ≤ S² / 4. So, the expression to minimize is x² + xS + P ≥ x² + xS + S² / 4. Because P is at most S² /4. So replacing P with the maximum possible (which would give the minimal expression?), wait, no. If we have P ≤ S² /4, then x² + xS + P ≤ x² + xS + S² /4. But we need to minimize the expression. Wait, if P can be smaller, then x² + xS + P can be smaller. But since the constraint is xP(x + S) = 1, if P is smaller, then x must be larger to keep the product xP(x + S) equal to 1. So maybe there is a balance here.Alternatively, maybe fix x and S, then P is determined by the constraint. Let's see. From the constraint: xP(x + S) = 1 => P = 1 / [x(x + S)]. Then, substituting back into the expression to minimize: x² + xS + P = x² + xS + 1 / [x(x + S)]. So now, the expression is in terms of x and S. Let's denote t = x + S. Then, x + S = t. But S = t - x. Wait, that might not help. Alternatively, perhaps set u = x(x + S). Then, the expression becomes x² + xS + 1/u. But u = x(x + S) = x² + xS. So the expression becomes u + 1/u. Therefore, the expression to minimize is u + 1/u where u = x² + xS. Since u is positive, the minimal value of u + 1/u is 2, achieved when u = 1. So, if we can set u = 1, then the expression would be minimized at 2. Therefore, the minimal value is 2.Wait, but is this possible? If u = 1, then x² + xS = 1. But from the constraint, P = 1 / [x(x + S)] = 1/u = 1. But P is yz, which must be positive. So, if P = 1, then yz = 1. Also, S = y + z. So, if yz = 1 and S = y + z, then by AM-GM, S ≥ 2√(yz) = 2. So, S ≥ 2. Also, x² + xS = 1. If S ≥ 2, then x² + 2x ≤ 1. But x² + 2x = (x + 1)^2 -1 ≤ 1 => (x + 1)^2 ≤ 2 => x + 1 ≤ sqrt(2) => x ≤ sqrt(2) -1 ≈ 0.414. So, possible. Let's see if there's a solution here.If u = 1, then x² + xS = 1, and P = 1. So, yz = 1, and S = y + z. Let me set y and z such that yz = 1, then S = y + 1/y (since z = 1/y). Then, substituting back into x² + xS = 1: x² + x(y + 1/y) = 1. Let me denote this equation as (1). Also, from the original constraint, xyz(x + y + z) = 1. But yz = 1, so x * 1 * (x + S) = 1 => x(x + S) = 1. But from u = x² + xS = 1, that's exactly the same. So, the condition u = 1 is equivalent to the constraint. Therefore, as long as we can find x, y, z such that yz = 1, S = y + z, and x(x + S) = 1, then this would satisfy the original constraint and give the minimal value of 2 for (x + y)(x + z).Wait, but let's check. For example, take y = z = 1. Then yz = 1, S = 2, then x(x + 2) = 1 => x² + 2x -1 = 0. Solving this quadratic equation: x = [-2 ± sqrt(4 + 4)] / 2 = [-2 ± sqrt(8)] / 2 = [-2 ± 2*sqrt(2)] / 2 = -1 ± sqrt(2). Since x is positive, x = sqrt(2) -1 ≈ 0.414. Then, (x + y)(x + z) = (sqrt(2) -1 +1)^2 = (sqrt(2))^2 = 2. So, that works. Alternatively, if we take different y and z such that yz = 1. Let's say y = 2, z = 1/2. Then S = y + z = 2 + 1/2 = 2.5. Then x(x + 2.5) = 1 => x² + 2.5x -1 =0. Solving: x = [-2.5 ± sqrt(6.25 +4)] / 2 = [-2.5 ± sqrt(10.25)] / 2 = [-2.5 ± 3.2]/2. Taking the positive solution: (0.7)/2 ≈ 0.35. Then, (x + y)(x + z) = (0.35 + 2)(0.35 + 0.5) = (2.35)(0.85) ≈ 2.0. Which is approximately 2.0. Wait, is that exactly 2? Let me compute:Suppose x ≈ 0.35, y = 2, z = 0.5. Then, (x + y)(x + z) = x² + x(z + y) + yz. We know that yz =1, and x(z + y) = x*2.5. From the equation x² + 2.5x =1, so (x + y)(x + z) = x² +2.5x +1 = (x² +2.5x) +1 =1 +1=2. So exactly 2. So regardless of y and z, as long as yz=1 and x is chosen accordingly, the expression is 2. Therefore, the minimal value is indeed 2. But wait, this seems too straightforward. Let me check if there's a case where (x + y)(x + z) could be less than 2. Suppose we don't set yz =1. Let's take some other values. For example, take x=1, y=1, z=1. Then, check the constraint: xyz(x + y + z) =1*1*1*(3)=3 ≠1. Not satisfying. So need to adjust. Let me find x, y, z that satisfy the constraint but aren't in the case where yz=1.Suppose x=1, then we have y z (1 + y + z) =1. Let's set y=1, then z(1 +1 + z) = z(2 + z) =1. So z² +2z -1=0. Solving, z = [-2 ± sqrt(4 +4)] /2 = [-2 ± sqrt(8)] /2 = [-2 ± 2*sqrt(2)]/2= -1 ± sqrt(2). So positive solution is z= sqrt(2)-1 ≈0.414. Then, (x + y)(x + z) = (1 +1)(1 +0.414)=2 *1.414≈2.828, which is bigger than 2. So, higher than our previous result.Another example: take x=0.5, then y z (0.5 + y + z)=1. Let me set y=0.5, then z (0.5 +0.5 + z)= z(1 + z)=1. So z² + z -1=0. Solution: z=(-1 ±sqrt(5))/2 ≈0.618. Then (x + y)(x + z)= (0.5 +0.5)(0.5 +0.618)=1*1.118≈1.118, but wait, that can't be. Wait, hold on. If x=0.5 and y=0.5, then z is approximately 0.618. So (x+y)=1, (x+z)=1.118, so the product is≈1.118, which is less than 2. But this contradicts our previous conclusion. But wait, let me check if the constraint is satisfied.Wait, x=0.5, y=0.5, z≈0.618. Then xyz(x + y + z)=0.5*0.5*0.618*(0.5+0.5+0.618)=0.5*0.5*0.618*(1.618)=0.25*0.618*1.618≈0.25*1≈0.25. But it should be 1. So this does not satisfy the constraint. Therefore, my mistake. I thought that by solving z(1 + z)=1, we get z≈0.618, but then the product xyz(x+y+z) is not 1. So this approach is invalid. Therefore, my initial thought that if we fix x and y and solve for z may not work because the product xyz(x + y + z) has to be 1. So, in order to satisfy the constraint, variables are more tightly coupled. Therefore, perhaps the minimal value is indeed 2, as found earlier when assuming yz=1. Alternatively, let's consider using Lagrange multipliers. Since we have a function to minimize subject to a constraint. Let me set up the Lagrangian. Let f(x, y, z) = (x + y)(x + z) and the constraint g(x, y, z) = xyz(x + y + z) -1 =0. Then, the Lagrangian is L = f - λg. Then, we take partial derivatives and set them to zero.First, compute partial derivatives:df/dx = (x + z) + (x + y) = 2x + y + z.df/dy = (x + z) + 0 = x + z.df/dz = (x + y) + 0 = x + y.Now, the partial derivatives of g:dg/dx = yz(x + y + z) + xyz(1) = yz(x + y + z) + xyz = yz(x + y + z + x).Wait, let me compute this carefully. g = xyz(x + y + z). So, derivative with respect to x:dg/dx = yz(x + y + z) + xyz*(1) = yz(x + y + z) + xyz = yz(x + y + z + x).Wait, that seems off. Let's compute it step by step. Let me write g = x y z (x + y + z). So, using product rule:dg/dx = y z (x + y + z) + x y z *1 = y z (x + y + z) + x y z.Similarly, dg/dy = x z (x + y + z) + x y z *1 = x z (x + y + z) + x y z.dg/dz = x y (x + y + z) + x y z *1 = x y (x + y + z) + x y z.Therefore, the partial derivatives are:dg/dx = y z (x + y + z + x) = y z (2x + y + z)Wait, no. Wait, original expression: dg/dx = y z (x + y + z) + x y z. Factor out y z:= y z [ (x + y + z) + x ] = y z (2x + y + z). Similarly:dg/dy = x z [ (x + y + z) + y ] = x z (x + 2y + z)dg/dz = x y [ (x + y + z) + z ] = x y (x + y + 2z)So, setting up the Lagrangian conditions:df/dx - λ dg/dx = 0 => 2x + y + z - λ y z (2x + y + z) = 0df/dy - λ dg/dy = 0 => x + z - λ x z (x + 2y + z) = 0df/dz - λ dg/dz = 0 => x + y - λ x y (x + y + 2z) = 0And the constraint: x y z (x + y + z) =1This system of equations seems complex, but maybe by assuming symmetry in y and z, we can simplify. Let's assume that y = z. Then, let's see what the equations become.Let y = z. Then, variables are x, y, y. The equations:1. 2x + 2y - λ y² (2x + 2y) = 02. x + y - λ x y (x + 2y + y) = x + y - λ x y (x + 3y) = 03. x + y - λ x y (x + y + 2y) = x + y - λ x y (x + 3y) = 0So equations 2 and 3 become the same, which is consistent with y = z. Now, let's process equation 1:Equation 1: 2x + 2y - λ y² (2x + 2y) = 0. Factor out 2:2(x + y) - λ y² * 2(x + y) = 0 => 2(x + y)(1 - λ y²) =0. Since x and y are positive, x + y ≠0, so 1 - λ y²=0 => λ =1/y².Equation 2: x + y - λ x y (x + 3y) =0. Substitute λ =1/y²:x + y - (1/y²) x y (x + 3y) =0 => x + y - (x / y)(x + 3y) =0.Multiply through by y to eliminate denominator:y(x + y) - x(x + 3y) =0 => xy + y² -x² -3xy =0 => -x² -2xy + y² =0 => x² + 2xy - y² =0.This is a quadratic in x: x² +2xy - y² =0. Solving for x:x = [-2y ± sqrt(4y² +4y²)] /2 = [-2y ± sqrt(8y²)]/2 = [-2y ± 2y√2]/2 = (-y ± y√2). Since x is positive, take the positive root:x = y(-1 + √2). So x = y(√2 -1).So x = (√2 -1)y. Then, let's use the constraint to find y.The constraint is xyz(x + y + z) =1. Since y = z and x = (√2 -1)y, substitute:x y² (x + 2y) =1 => (√2 -1)y * y² * ( (√2 -1)y + 2y ) =1.Simplify the expression inside the parenthesis:(√2 -1)y + 2y = [ (√2 -1) +2 ] y = (√2 +1)y.So, substituting back:(√2 -1)y * y² * (√2 +1)y = (√2 -1)(√2 +1) y^4 = (2 -1)y^4 = y^4 =1.Therefore, y^4=1 => y=1 (since y>0). Therefore, y=1, z=1, x=√2 -1. Which matches the earlier solution. Therefore, the critical point is at x=√2 -1, y=z=1. Then, (x + y)(x + z)=(√2 -1 +1)^2=(√2)^2=2. Therefore, by the method of Lagrange multipliers, we also find that the minimal value is 2, achieved at x=√2 -1, y=z=1. Now, to ensure that this is indeed a minimum, we should check the second derivative or the bordered Hessian, but given that the problem is set in positive reals and we have only one critical point from the symmetry assumption, and since we tested another case where (x + y)(x + z) was larger, it's reasonable to conclude that this is the global minimum.Another approach: Let's use substitution. Let me set a = x + y and b = x + z. Then, we need to minimize ab. Let's express the constraint in terms of a and b.But x, y, z are positive, so y = a - x, z = b - x. Then, the constraint becomes x(a - x)(b - x)(x + (a - x) + (b - x)) =1.Simplify the sum inside: x + a -x + b -x = a + b -x. So the constraint is x(a - x)(b - x)(a + b -x) =1.This seems complicated. Maybe not the best approach. Alternatively, note that (x + y)(x + z) =x(x + y + z) + yz. Let's denote S =x + y + z. Then, the expression becomes xS + yz. The constraint is xyzS=1. So, we can write yz=1/(xS). Therefore, the expression becomes xS + 1/(xS). Let me set t =xS. Then, the expression is t +1/t, which again has a minimum of 2 when t=1. Therefore, the minimal value is 2, achieved when t=1, i.e., xS=1. But from the constraint, xyzS=1. If xS=1, then y z=1. Therefore, if we can achieve xS=1 and yz=1, then both the constraint and the expression are satisfied. Therefore, the minimal value is 2. Therefore, regardless of the method used—whether assuming symmetry, Lagrange multipliers, or algebraic manipulation—the minimal value is 2. To confirm, let's check another point. Suppose x=1, y=1, then the constraint requires z(1 +1 + z)=1/(1*1*z) => z(2 + z) =1/(z). Multiply both sides by z: z²(2 + z)=1. So, 2z² + z³ -1=0. Let me try z=0.5: 2*(0.25)+0.125 -1=0.5 +0.125 -1= -0.375. z=0.6: 2*0.36 +0.216 -1=0.72 +0.216 -1= -0.064. z=0.7: 2*0.49 +0.343 -1=0.98 +0.343 -1=0.323. So between 0.6 and 0.7. Let me try z=0.63: 2*(0.3969) +0.250047 -1≈0.7938 +0.250047 -1≈1.0438 -1≈0.0438. Close. z=0.62: 2*(0.3844) +0.238328 -1≈0.7688 +0.2383≈1.0071 -1≈0.0071. z=0.615: 2*(0.615²)=2*(0.378225)=0.75645 +0.615³≈0.615*0.615*0.615≈0.615*0.615=0.378225*0.615≈0.2326. Total≈0.75645+0.2326≈0.989. So≈0.989 -1≈-0.011. So between 0.615 and 0.62. Therefore, solution around z≈0.617. Then, (x + y)(x + z)= (1 +1)(1 +0.617)=2*1.617≈3.234>2. So, again, higher than 2. Therefore, all these checks suggest that 2 is indeed the minimal value. Thus, the answer is boxed{2}.Final Answerboxed{2}</think>Given positive real numbers (x, y, z) that satisfy the equation (xyz(x + y + z) = 1), we need to find the minimum value of ((x + y)(x + z)).First, we note that the expression ((x + y)(x + z)) can be expanded as (x^2 + xz + xy + yz). We aim to minimize this expression under the given constraint. By assuming symmetry in (y) and (z), we set (y = z). This simplifies the constraint to (x y^2 (x + 2y) = 1) and the expression to minimize becomes ((x + y)^2). Let (t = x + y), then substituting (x = t - y) into the constraint gives ((t^2 - y^2)y^2 = 1). Solving this, we find that the minimum value of (t^2) is 2 when (y = 1) and (x = sqrt{2} - 1), leading to ((x + y)^2 = 2).To confirm this result without assuming symmetry, we use substitution and the AM-GM inequality. We express the constraint as (xyz(x + y + z) = 1) and rewrite the expression ((x + y)(x + z)) in terms of (x), (y), and (z). By setting (u = x(x + y + z)), we find that the expression to minimize becomes (u + frac{1}{u}), which has a minimum value of 2 when (u = 1).Using Lagrange multipliers, we verify that the critical point occurs when (x = sqrt{2} - 1), (y = z = 1), leading to the same minimum value of 2.Thus, the minimum value of ((x + y)(x + z)) is (boxed{2}).,<think>Okay, let's see. I need to find the minimum value of (x + y)(x + z) given that x, y, z are positive real numbers and they satisfy the equation xyz(x + y + z) = 1. Hmm, okay. Alright, first, let me understand what's given and what's being asked.We have three variables: x, y, z, all positive real numbers. The condition is xyz(x + y + z) = 1. And we need to minimize the expression (x + y)(x + z). Hmm, so maybe I can express this in terms of variables that can be minimized using some inequality? Like AM-GM or Cauchy-Schwarz? Let me think. Let me first try expanding (x + y)(x + z) to see if that helps.Expanding (x + y)(x + z) gives x² + xz + xy + yz. So that's x² + x(y + z) + yz. Hmm, but how does this relate to the given condition? The given condition has xyz times (x + y + z). Maybe I can find a relationship between these terms.Alternatively, perhaps substitution might help. Let me see if I can express some variables in terms of others. For example, maybe set t = x + y + z. Then the condition becomes xyzt = 1. But I don't know if that helps directly. Wait, but if t = x + y + z, then (x + y)(x + z) = x² + x(y + z) + yz. Since t = x + y + z, then y + z = t - x. So substituting back, we have x² + x(t - x) + yz = x² + xt - x² + yz = xt + yz. So (x + y)(x + z) simplifies to xt + yz. Interesting. So (x + y)(x + z) = x(x + y + z) + yz. Hmm, that's a nice simplification. Let me note that.So, (x + y)(x + z) = x(x + y + z) + yz. Let's denote S = x + y + z. Then the expression becomes xS + yz. Given that xyzS = 1. So maybe I can express yz in terms of x and S. Let's see.From xyzS = 1, we can solve for yz: yz = 1/(xS). So substituting back into the expression, we have xS + 1/(xS). So the expression to minimize is xS + 1/(xS). Hmm, that's interesting. So now, if I can express this in terms of a single variable or find a way to apply AM-GM, maybe?But S is x + y + z, which is related to x, y, z. However, y and z are variables here. So perhaps I need to consider how S relates to x. Wait, if I can fix x and S, then yz is determined by yz = 1/(xS). But given that S = x + y + z, and y and z are positive, then y + z = S - x, and yz = 1/(xS). So given y + z and yz, we can think of y and z as roots of the quadratic equation t² - (S - x)t + 1/(xS) = 0. For real positive roots, the discriminant must be non-negative. So discriminant D = (S - x)² - 4/(xS) ≥ 0. Hmm, maybe this is a way to proceed? Let me check.So discriminant D = (S - x)² - 4/(xS) ≥ 0. So (S - x)² ≥ 4/(xS). That's a relationship between S and x. But S is x + y + z, which complicates things. Alternatively, maybe there's a better approach. Let me think.Alternatively, maybe use the AM-GM inequality on the expression xS + 1/(xS). Let's let’s denote A = xS. Then the expression becomes A + 1/A. The function f(A) = A + 1/A has a minimum at A = 1, where f(A) = 2. But wait, is A a variable here? But A = xS, and S is dependent on x. So even if I consider A as xS, maybe I can use AM-GM here.But since S = x + y + z, and y and z are positive, then S > x. Therefore, A = xS > x². But perhaps if we can show that xS ≥ something, then maybe. Alternatively, perhaps not. Let me check. Wait, but if we could set A = xS = 1, then xS = 1, then the expression would be 1 + 1/1 = 2. So if we can have xS = 1, then the minimum would be 2. But is xS = 1 possible?If xS = 1, then since S = x + y + z, we have x(x + y + z) = 1. But also, the given condition is xyzS = 1. So substituting S = 1/x into xyzS, we get xyz*(1/x) = yz = 1. So yz = 1. Then, we have S = x + y + z = 1/x. So x + y + z = 1/x, and yz = 1.So then, we can consider that given yz = 1 and S = 1/x, then y + z = S - x = (1/x) - x. So y and z satisfy y + z = (1/x) - x and yz = 1. Then, the quadratic equation would be t² - [(1/x) - x]t + 1 = 0. The discriminant here is [(1/x) - x]² - 4*1*1 = (1/x² - 2 + x²) - 4 = 1/x² + x² - 6.For real roots, discriminant must be non-negative. So 1/x² + x² - 6 ≥ 0. Let’s set u = x². Then 1/u + u - 6 ≥ 0. Multiply both sides by u (since u > 0): 1 + u² - 6u ≥ 0 → u² - 6u + 1 ≥ 0. Solving quadratic inequality: u = [6 ± sqrt(36 - 4)]/2 = [6 ± sqrt(32)]/2 = [6 ± 4*sqrt(2)]/2 = 3 ± 2*sqrt(2). So u ≥ 3 + 2√2 or u ≤ 3 - 2√2. But u = x² must be positive, and 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172, which is positive. Therefore, x² ≤ 3 - 2√2 or x² ≥ 3 + 2√2. However, since x is positive, x ≤ sqrt(3 - 2√2) or x ≥ sqrt(3 + 2√2). Let me compute sqrt(3 - 2√2). Let's see, sqrt(3 - 2√2) is sqrt((√2 - 1)^2) = √2 - 1 ≈ 0.414. Similarly, sqrt(3 + 2√2) is sqrt((√2 + 1)^2) = √2 + 1 ≈ 2.414.But if x ≥ sqrt(3 + 2√2) ≈ 2.414, then y + z = (1/x) - x. Wait, but if x is large, (1/x) - x becomes negative, which is impossible because y and z are positive. Therefore, x cannot be too large. Hence, x must be ≤ sqrt(3 - 2√2) ≈ 0.414. Wait, but x is positive. So in this case, when xS = 1, we need that (1/x) - x > 0, so 1/x > x → x² < 1 → x < 1. Which is true for x ≤ sqrt(3 - 2√2) ≈ 0.414, which is less than 1. So in that case, y + z = (1/x) - x is positive.Therefore, the discriminant condition gives us possible x values. However, even if we can have real y and z, we need to check if this leads to the minimal value. If we set A = xS = 1, then the expression is 2, but we need to verify if such x, y, z exist. Since for x ≤ sqrt(3 - 2√2), discriminant is non-negative, so there exist real y, z. But since we need y and z to be positive, we need the roots of the quadratic equation to be positive. Let's check.The quadratic equation is t² - [(1/x) - x]t + 1 = 0. The sum of roots is (1/x) - x, and the product is 1. So if the sum of roots is positive and the product is positive, then both roots are positive. Since product is 1, positive. Sum is (1/x) - x. If x < 1, then (1/x) - x is positive. So yes, as x is positive and less than 1, sum is positive. Therefore, y and z are positive. Therefore, such y and z exist.Therefore, if we can set xS = 1, then (x + y)(x + z) = 2. But is this achievable? Let's check with specific numbers. Let me take x = sqrt(3 - 2√2). Let's compute that.sqrt(3 - 2√2) = sqrt((√2 - 1)^2) = √2 - 1. So x = √2 - 1 ≈ 0.414. Then S = 1/x = 1/(√2 - 1) = √2 + 1 (rationalizing the denominator). Then y + z = S - x = (√2 + 1) - (√2 - 1) = 2. And yz = 1. Therefore, y and z are roots of t² - 2t + 1 = 0, which factors as (t - 1)^2 = 0. Therefore, y = z = 1. So in this case, x = √2 - 1, y = z = 1. Let's check if this satisfies the original condition.Compute xyzS: x = √2 - 1, y = 1, z = 1, S = x + y + z = (√2 - 1) + 1 + 1 = √2 + 1. Then xyzS = (√2 - 1)*1*1*(√2 + 1). Multiply (√2 - 1)(√2 + 1) = (√2)^2 - 1^2 = 2 - 1 = 1. Therefore, xyzS = 1, which satisfies the condition. And (x + y)(x + z) = ( (√2 - 1) + 1 )^2 = (√2)^2 = 2. So yes, this gives us the value 2, and it satisfies the conditions.So if we can achieve the value 2, and since the expression (x + y)(x + z) is equal to xS + yz = xS + 1/(xS), and the minimum of A + 1/A is 2 when A = 1, then this must be the minimal value. Therefore, the minimal value is 2. But let me check if there are any other cases where it could be lower. Suppose, for instance, that xS is not equal to 1. Then, since the function A + 1/A is minimized at A = 1, any deviation from A = 1 would lead to a larger value. Therefore, the minimal value is indeed 2. So the answer should be 2, boxed.Wait, but let me check another example to be sure. Let's take x = 1, then S = x + y + z = 1 + y + z. The condition is xyzS = 1* y* z*(1 + y + z) = 1. Let's set y = z for simplicity. Then y = z. Then, we have y²(1 + 2y) = 1. Let's solve for y: y²(1 + 2y) = 1. Let’s set t = y, then t² + 2t³ = 1. Hmm, 2t³ + t² - 1 = 0. Let's see if t=0.5: 2*(0.125) + 0.25 -1 = 0.25 + 0.25 -1 = -0.5. t=0.6: 2*(0.216) + 0.36 -1 ≈ 0.432 + 0.36 -1 ≈ -0.208. t=0.7: 2*(0.343) + 0.49 -1 ≈ 0.686 + 0.49 -1 ≈ 0.176. So between t=0.6 and t=0.7. Let's approximate. Maybe t≈0.65: 2*(0.2746) + 0.4225 -1 ≈ 0.5492 + 0.4225 -1 ≈ -0.0283. Close to zero. t=0.66: 2*(0.2875) + 0.4356 -1 ≈ 0.575 + 0.4356 -1 ≈ 0.0106. So solution around t≈0.66. Let's take t≈0.66. Then y≈0.66, z≈0.66. Then S≈1 + 0.66 + 0.66 ≈2.32. Then (x + y)(x + z) = (1 + 0.66)^2 ≈2.7556. Which is greater than 2. So this case gives a larger value. Therefore, the minimal value is indeed 2. Hence, I think the answer is 2.Final Answerboxed{2}</think>Given ( x, y, z in mathbf{R}^+ ) satisfying ( xyz(x + y + z) = 1 ), we need to find the minimum value of ( (x + y)(x + z) ).First, we expand ( (x + y)(x + z) ):[(x + y)(x + z) = x^2 + xz + xy + yz = x^2 + x(y + z) + yz]Let ( S = x + y + z ). Then, we can rewrite the expression as:[(x + y)(x + z) = xS + yz]Given the condition ( xyzS = 1 ), we solve for ( yz ):[yz = frac{1}{xS}]Substituting this back into the expression, we get:[(x + y)(x + z) = xS + frac{1}{xS}]Let ( A = xS ). The expression becomes ( A + frac{1}{A} ), which is minimized when ( A = 1 ) by the AM-GM inequality. Thus, the minimum value is:[A + frac{1}{A} geq 2]We check if ( A = 1 ) is achievable. If ( xS = 1 ), then:[xyzS = 1 implies yz = 1]Given ( S = x + y + z ), we have ( y + z = S - x ). Since ( yz = 1 ), the quadratic equation ( t^2 - (S - x)t + 1 = 0 ) must have real positive roots. The discriminant must be non-negative:[(S - x)^2 - 4 geq 0]Substituting ( S = frac{1}{x} ), we get:[left( frac{1}{x} - x right)^2 - 4 geq 0]Solving this, we find ( x = sqrt{2} - 1 ), ( y = z = 1 ), which satisfies the condition ( xyzS = 1 ) and gives:[(x + y)(x + z) = (sqrt{2} - 1 + 1)^2 = (sqrt{2})^2 = 2]Thus, the minimum value of ( (x + y)(x + z) ) is (boxed{2}).]

🔑:11. Given x^{2}+x(y+z)=frac{1}{y z}, thus(x+y)(x+z)=left[x^{2}+x(y+z)right]+y z=frac{1}{y z}+y z geqslant 2 sqrt{frac{1}{y z} cdot y z}=2.The equality holds if and only if y z=1.Therefore, the minimum value of (x+y)(x+z) is 2.

🧠:[<think>Okay, let's tackle these two problems one by one. Starting with the first one about the association between height and lung capacity levels. So, we have a contingency table here. The rows are height categories: below 175 cm and 175 cm or above. The columns are lung capacity levels, but wait, the table seems a bit confusing. Let me parse it again. The columns are labeled "Lung Capacity Level" and "Total", but the entries under "Lung Capacity Level" are numbers. Maybe that's two categories? Wait, looking at the totals, the first row total is 44, and under "Lung Capacity Level" there's 22 and 22. Similarly, the second row has 30 and 6 summing to 36. The total column has 52 and 28. Wait, maybe the "Lung Capacity Level" is split into two categories? But the original problem mentions four levels: excellent, good, qualified, unqualified. Hmm, this is conflicting. Wait, hold on. Let me check again. The problem says the contingency table shows lung capacity levels (excellent, good, qualified, unqualified) of 80 male students. But in the table, under "Lung Capacity Level", there are two numbers: 22 and 22 for the first row, 30 and 6 for the second row. The total column shows 52 and 28. So maybe the columns are actually two categories instead of four? That seems contradictory. Perhaps there's a formatting issue here. Maybe the columns are supposed to be two categories, but the original problem mentions four. Hmm. Wait, maybe "excellent, good, qualified, unqualified" are the rows or columns? Let me read the problem again."A certain school conducts physical fitness tests... The lung capacity levels (excellent, good, qualified, unqualified) of 80 male students... The results are shown in the following contingency table:| Height | Lung Capacity Level | Total ||--------|----------------------|-------|| Below 175 cm | 22 | 22 | 44 || 175 cm or above | 30 | 6 | 36 || Total | 52 | 28 | 80 |"Wait a second, the table seems to have two columns under "Lung Capacity Level". But the problem mentions four levels. Maybe the columns are collapsed into two categories? For example, maybe "excellent and good" vs. "qualified and unqualified"? The problem doesn't specify. That could be a problem. Alternatively, maybe the table is misformatted. Let me check the original problem statement again.Wait, the user wrote:| Height | Lung Capacity Level | Total ||--------|----------------------|-------|| Below 175 cm | 22 | 22 | 44 || 175 cm or above | 30 | 6 | 36 || Total | 52 | 28 | 80 |So, "Lung Capacity Level" is a single column with two entries per row? That can't be. Maybe "Lung Capacity Level" is split into two categories. Maybe the columns are two levels, but the problem mentions four. This is confusing. Wait, perhaps the original data had four levels, but in the contingency table, they combined them into two categories? For instance, excellent/good vs. qualified/unqualified. The problem doesn't specify, but given that the totals add up (22+22=44, 30+6=36, 52+28=80), it's likely that "Lung Capacity Level" is split into two categories here. Since the problem mentions four levels, but the table only has two numbers per row, maybe there was a mistake in the problem's translation or formatting. Alternatively, perhaps "Lung Capacity Level" is the header for two columns, but it's written as one. Let me try to visualize the table properly. Maybe the correct table should have two columns for lung capacity: maybe "High" (excellent and good) and "Low" (qualified and unqualified). Then, the entries would be 22 and 22 for below 175 cm, 30 and 6 for above 175 cm. Then the totals would be 52 and 28. So, total high lung capacity is 52, low is 28. So, 52+28=80. That makes sense. The problem mentions four levels, but perhaps they grouped them into two for the contingency table. Since the question is about association between height and lung capacity levels, maybe the two categories are sufficient for the chi-squared test. Assuming that, then we can proceed with a 2x2 contingency table. The question is whether there's an association between height and lung capacity levels at 99.5% confidence. That corresponds to a significance level α = 0.005 (since 1 - 0.995 = 0.005). Given that, we need to perform a chi-squared test for independence. The formula provided is K^2 = n(ad - bc)^2 / [(a+b)(c+d)(a+c)(b+d)], which is the formula for the chi-squared statistic in a 2x2 table. Here, a, b, c, d are the cells of the table. Let me set up the table properly.Let me reconstruct the contingency table. Let's assume that "Lung Capacity Level" has two categories, say "High" and "Low". Then:| Height | High | Low | Total ||-----------------|------|-----|-------|| Below 175 cm | 22 | 22 | 44 || 175 cm or above | 30 | 6 | 36 || Total | 52 | 28 | 80 |So, a = 22, b = 22, c = 30, d = 6. Then, n = a + b + c + d = 80.Plugging into the formula:K^2 = 80*(22*6 - 22*30)^2 / [(22+22)(30+6)(22+30)(22+6)]First compute ad - bc: 22*6 - 22*30 = 132 - 660 = -528. Then square it: (-528)^2 = 278,784.Multiply by n: 80 * 278,784 = 22,302,720.Denominator: (44)(36)(52)(28). Let's compute each part:44 * 36 = 1,58452 * 28 = 1,456Then multiply those two: 1,584 * 1,456. Let's compute this:First, 1,500 * 1,456 = 2,184,000Then 84 * 1,456 = 122, 304 (Wait: 80*1,456=116,480 and 4*1,456=5,824; 116,480 + 5,824 = 122,304)So total denominator: 2,184,000 + 122,304 = 2,306,304Therefore, K^2 = 22,302,720 / 2,306,304 ≈ 9.674Now, we need to compare this value to the critical value from the chi-squared distribution table. The degrees of freedom for a 2x2 contingency table is (rows - 1)(columns - 1) = 1*1 = 1. The given table for K^2 has critical values at different significance levels. For 99.5% confidence, the significance level α is 0.005, so we look at the 0.005 column. The critical value is 7.879. Our calculated K^2 is approximately 9.674, which is greater than 7.879. Therefore, we can reject the null hypothesis and conclude that there is an association between height and lung capacity levels at the 99.5% confidence level.Wait, but wait. Let me check my calculations again to be sure.Calculating ad - bc: 22*6 = 132, 22*30 = 660. 132 - 660 = -528. Squared is 278,784. Multiply by 80: 278,784 * 80. Let's compute 278,784 * 80. 278,784 * 10 = 2,787,840. Multiply by 8: 22,302,720. Correct.Denominator: (a+b)(c+d)(a+c)(b+d). So (22+22)=44, (30+6)=36, (22+30)=52, (22+6)=28. Multiplying 44*36=1,584 and 52*28=1,456. Then 1,584 * 1,456. Let me check that multiplication again.1,584 * 1,456:Break it down:1,584 * 1,000 = 1,584,0001,584 * 400 = 633,6001,584 * 50 = 79,2001,584 * 6 = 9,504Add them up:1,584,000 + 633,600 = 2,217,6002,217,600 + 79,200 = 2,296,8002,296,800 + 9,504 = 2,306,304. Correct.So K^2 = 22,302,720 / 2,306,304 ≈ 9.674The critical value at α = 0.005 (which is 99.5% confidence) for df=1 is 7.879. Since 9.674 > 7.879, we reject the null hypothesis. Therefore, there is a significant association between height and lung capacity levels at the 99.5% confidence level. So the answer to part (1) is yes.Moving on to part (2). The problem states: A fitness test group consists of 6 male students, including 1 unqualified, 4 good, and 1 excellent lung capacity levels. The scoring system: unqualified - 0 points, qualified - 1 point, good - 2 points, excellent - 3 points. Two students are randomly selected. Let X be the sum of their scores. Find the probability distribution and the mean of X.First, let's clarify the composition of the group. It says 6 male students: 1 unqualified, 4 good, 1 excellent. Wait, but the scoring system includes qualified as 1 point, but there's no mention of qualified students in the group. The group has 1 unqualified (0), 4 good (2), and 1 excellent (3). So total students: 1 + 4 + 1 = 6. That's correct.So, the students are:- 1 student with score 0 (unqualified)- 4 students with score 2 (good)- 1 student with score 3 (excellent)We need to select two students at random and find the sum of their scores, X. We have to list all possible pairs, compute their sums, find the probabilities, and then compute the expected value.First, let's determine the total number of ways to choose 2 students out of 6. That's C(6,2) = 15. So there are 15 possible pairs.Now, let's list all possible pairs and their scores:Possible pairs:1. 0 and 22. 0 and 2 (another 2)3. 0 and 2 (another 2)4. 0 and 2 (another 2)5. 0 and 36. 2 and 27. 2 and 28. 2 and 29. 2 and 210. 2 and 211. 2 and 212. 2 and 313. 2 and 314. 2 and 315. 2 and 3Wait, but this is not the correct way to list them. Let's instead consider the different combinations based on the categories.We have:- 1 student with score 0 (U)- 4 students with score 2 (G1, G2, G3, G4)- 1 student with score 3 (E)So, the possible pairs are:1. U and G1: sum = 0 + 2 = 22. U and G2: sum = 23. U and G3: sum = 24. U and G4: sum = 25. U and E: sum = 0 + 3 = 36. G1 and G2: sum = 2 + 2 = 47. G1 and G3: sum = 48. G1 and G4: sum = 49. G1 and E: sum = 2 + 3 = 510. G2 and G3: sum = 411. G2 and G4: sum = 412. G2 and E: sum = 513. G3 and G4: sum = 414. G3 and E: sum = 515. G4 and E: sum = 5Wait, actually, the pairs can be categorized as follows:- Pairs involving U and G: There are 4 such pairs (U with each of the 4 Gs). Each has a sum of 2.- Pair involving U and E: 1 pair, sum 3.- Pairs involving two Gs: C(4,2) = 6 pairs. Each has a sum of 4.- Pairs involving G and E: 4 pairs (each G with E). Each has a sum of 5.Total pairs: 4 + 1 + 6 + 4 = 15. Correct.So, the possible sums X can take are 2, 3, 4, 5.Now, let's count the number of pairs for each sum:- Sum = 2: 4 pairs.- Sum = 3: 1 pair.- Sum = 4: 6 pairs.- Sum = 5: 4 pairs.Therefore, the probability distribution is:P(X=2) = 4/15P(X=3) = 1/15P(X=4) = 6/15 = 2/5P(X=5) = 4/15To write the probability distribution, we list each possible value of X with its probability.Now, the mean (expected value) of X is calculated as E[X] = Σ [x * P(X=x)].Calculating each term:For X=2: 2 * (4/15) = 8/15For X=3: 3 * (1/15) = 3/15 = 1/5For X=4: 4 * (6/15) = 24/15 = 8/5For X=5: 5 * (4/15) = 20/15 = 4/3Adding these up:8/15 + 1/5 + 8/5 + 4/3Convert all to fifteenths:8/15 + 3/15 + 24/15 + 20/15 = (8 + 3 + 24 + 20)/15 = 55/15 = 11/3 ≈ 3.6667Alternatively, to check, maybe there's another way to compute the expected value.Alternative approach: The expected value of the sum is the sum of the expected values of each individual score, but since we are sampling without replacement, the covariance needs to be considered. Wait, but actually, linearity of expectation holds regardless of independence. So, E[X1 + X2] = E[X1] + E[X2], where X1 and X2 are the scores of the two selected students. Since the selection is without replacement, the expectation can still be calculated as 2 * E[X1], because the two are identically distributed.First, compute the expected value of a single student's score.There are 6 students:1 student: 04 students: 21 student: 3Total scores: 0 + 4*2 + 1*3 = 0 + 8 + 3 = 11Mean score per student: 11/6 ≈ 1.8333Therefore, the expected sum for two students is 2*(11/6) = 22/6 = 11/3 ≈ 3.6667, which matches our previous result. So that's a good check.Therefore, the mean is 11/3.So, putting it all together, the probability distribution is:X | P(X)---|---2 | 4/153 | 1/154 | 6/155 | 4/15And the mean is 11/3.But let me check the calculation again for the probability distribution. The counts:- Sum 2: 4 pairs- Sum 3: 1 pair- Sum 4: 6 pairs- Sum 5: 4 pairsTotal pairs: 15. So the probabilities are correct.Yes, so that's the distribution. To present the probability distribution, we can list each possible X with its probability. The mean is 11/3, which is approximately 3.6667.Final Answer(1) boxed{Yes}(2) The probability distribution of ( X ) is:[begin{array}{c|c}X & P(X) hline2 & dfrac{4}{15} 3 & dfrac{1}{15} 4 & dfrac{2}{5} 5 & dfrac{4}{15} end{array}]The mean of ( X ) is boxed{dfrac{11}{3}}.</think>To determine if there is an association between the height of male students and their lung capacity levels at a 99.5% confidence level, we perform a chi-squared test for independence. The contingency table is as follows:| Height | High | Low | Total ||-----------------|------|-----|-------|| Below 175 cm | 22 | 22 | 44 || 175 cm or above | 30 | 6 | 36 || Total | 52 | 28 | 80 |Using the formula for the chi-squared statistic ( K^2 ):[K^2 = frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}]where ( a = 22 ), ( b = 22 ), ( c = 30 ), ( d = 6 ), and ( n = 80 ):[K^2 = frac{80(22 cdot 6 - 22 cdot 30)^2}{(22+22)(30+6)(22+30)(22+6)} = frac{80(-528)^2}{44 cdot 36 cdot 52 cdot 28} approx 9.674]Comparing this value to the critical value from the chi-squared distribution table at ( alpha = 0.005 ) (99.5% confidence level) with 1 degree of freedom, the critical value is 7.879. Since ( 9.674 > 7.879 ), we reject the null hypothesis, concluding there is an association.For the probability distribution and mean of ( X ):The group consists of 1 unqualified (0 points), 4 good (2 points), and 1 excellent (3 points). The possible pairs and their sums are:- Sum 2: 4 pairs (0 and 2)- Sum 3: 1 pair (0 and 3)- Sum 4: 6 pairs (2 and 2)- Sum 5: 4 pairs (2 and 3)The probability distribution is:[begin{array}{c|c}X & P(X) hline2 & dfrac{4}{15} 3 & dfrac{1}{15} 4 & dfrac{2}{5} 5 & dfrac{4}{15} end{array}]The mean ( E[X] ) is calculated as:[E[X] = 2 cdot frac{4}{15} + 3 cdot frac{1}{15} + 4 cdot frac{6}{15} + 5 cdot frac{4}{15} = frac{8 + 3 + 24 + 20}{15} = frac{55}{15} = frac{11}{3}] Final Answer(1) boxed{Yes}(2) The probability distribution of ( X ):[begin{array}{c|c}X & P(X) hline2 & dfrac{4}{15} 3 & dfrac{1}{15} 4 & dfrac{2}{5} 5 & dfrac{4}{15} end{array}]The mean of ( X ) is boxed{dfrac{11}{3}}.,<think>Okay, so I have two questions here about statistics and probability. Let me tackle them one by one. Starting with question (1). The problem is asking whether we can be 99.5% confident that there's an association between the height of male students and their lung capacity levels. They provided a contingency table and the formula for K-squared (which I think is the chi-squared statistic). The table has the observed frequencies, and we need to perform a chi-squared test of independence. First, let me recall that a chi-squared test is used to determine if there's a significant association between two categorical variables. Here, the two variables are height (below 175 cm or 175 cm and above) and lung capacity level (excellent, good, qualified, unqualified). Wait, but looking at the contingency table they provided, it seems like lung capacity levels are split into two categories instead of four. Wait, the table has "Lung Capacity Level" with two entries per row: the first row has 22 and 22, and the second row has 30 and 6. Then the totals are 52 and 28. Hmm, maybe there's a typo in the problem? Because the description mentions four levels (excellent, good, qualified, unqualified), but the contingency table only has two columns. Let me check again.The problem states: "The lung capacity levels (excellent, good, qualified, unqualified) of 80 male students... results shown in the contingency table..." But the table has "Lung Capacity Level" with two columns. Wait, maybe the columns are combined? Like, perhaps the first column is "Excellent/Good" and the second is "Qualified/Unqualified"? Or maybe it's a 2x2 table where the columns are two categories instead of four. The problem might have simplified the table for the sake of the question. Let me check the given formula for K-squared, which is the chi-squared statistic formula for a 2x2 contingency table. The formula is given as K² = n(ad - bc)² / [(a+b)(c+d)(a+c)(b+d)], which is indeed the formula for the chi-squared statistic for a 2x2 table. So the contingency table provided is a 2x2 table. Therefore, the lung capacity levels must have been categorized into two groups instead of four. Maybe the original four levels were collapsed into two, like "Pass/Fail" or "High/Low". The problem statement might have a mistake, but given the table, we can work with that.So the contingency table is:Height | Lung Capacity Level | Total----------------|---------------------|-------Below 175 cm | 22 | 22 | 44175 cm or above | 30 | 6 | 36Total | 52 | 28 | 80So the rows are height categories, and the columns are lung capacity levels, which have been grouped into two categories. Let's denote the first column as, say, "High" (excellent and good) and the second as "Low" (qualified and unqualified), but the exact labels don't matter for the calculation. The key is that it's a 2x2 table.To perform the chi-squared test, we need to compute the chi-squared statistic using the given formula and compare it to the critical value at the 99.5% confidence level. The confidence level corresponds to the significance level alpha. Since 99.5% confidence means alpha is 0.005 (1 - 0.995 = 0.005). From the table provided, for alpha=0.005, the critical value k is 7.879. So if our calculated chi-squared statistic is greater than or equal to 7.879, we reject the null hypothesis (that there's no association) and conclude there is a significant association.Let me first compute the chi-squared statistic using the formula given. The formula is K² = [n(ad - bc)²] / [(a+b)(c+d)(a+c)(b+d)], where a, b, c, d are the cells of the contingency table.Let's set up the table:For the first row (Below 175 cm):a = 22 (first column)b = 22 (second column)For the second row (175 cm or above):c = 30 (first column)d = 6 (second column)Total n = 80.So plugging into the formula:First, compute (ad - bc):ad = 22 * 6 = 132bc = 22 * 30 = 660ad - bc = 132 - 660 = -528Then square that: (-528)² = 278,784Multiply by n: 80 * 278,784 = 22,302,720Now the denominator: (a+b)(c+d)(a+c)(b+d)Compute each term:(a + b) = 22 + 22 = 44(c + d) = 30 + 6 = 36(a + c) = 22 + 30 = 52(b + d) = 22 + 6 = 28Multiply them all together: 44 * 36 * 52 * 28Let me compute step by step:First, 44 * 36 = 1,584Then, 52 * 28 = 1,456Now multiply 1,584 * 1,456. Hmm, that's a big number. Let me compute that:1,584 * 1,456:Break it down:1,584 * 1,000 = 1,584,0001,584 * 400 = 633,6001,584 * 50 = 79,2001,584 * 6 = 9,504Adding up:1,584,000 + 633,600 = 2,217,6002,217,600 + 79,200 = 2,296,8002,296,800 + 9,504 = 2,306,304So the denominator is 2,306,304.Therefore, K² = 22,302,720 / 2,306,304 ≈Let me do the division:Divide numerator and denominator by 1000: 22,302.72 / 2,306.304Approximately, 22,302.72 ÷ 2,306.304 ≈ Let's see:2,306.304 * 9 = 20,756.736Subtract that from 22,302.72: 22,302.72 - 20,756.736 = 1,545.984Now, 2,306.304 * 0.67 ≈ 2,306.304 * 0.6 = 1,383.7824; 2,306.304 * 0.07 = 161.44128; total ≈1,545.22368So 9.67. Therefore, approximately 9.67. Let me check with calculator steps:22,302,720 ÷ 2,306,304Divide numerator and denominator by 48:22,302,720 ÷ 48 = 464,6402,306,304 ÷ 48 = 48,048So now, 464,640 ÷ 48,048 ≈ Let's divide numerator and denominator by 48:464,640 ÷ 48 = 9,68048,048 ÷ 48 = 1,001So 9,680 ÷ 1,001 ≈ 9.67Yes, so approximately 9.67.So K² ≈ 9.67.Now, the critical value at alpha=0.005 is 7.879. Since 9.67 > 7.879, we reject the null hypothesis. Therefore, we can be 99.5% confident that there is an association between height and lung capacity level.Wait, but hold on. The formula given is K² = n(ad - bc)² / [(a+b)(c+d)(a+c)(b+d)]. This is the formula for the chi-squared statistic for a 2x2 contingency table. However, the chi-squared test usually requires that the expected frequencies are not too small. Let me check if any expected cell counts are less than 5. The expected frequency for each cell is (row total * column total) / n.So for the first cell (Below 175 cm and first lung capacity column):Expected = (44 * 52) / 80 = (2288) / 80 = 28.6Second cell (Below 175 cm and second column):Expected = (44 * 28) / 80 = 1232 / 80 = 15.4Third cell (175 cm or above and first column):Expected = (36 * 52) / 80 = 1872 / 80 = 23.4Fourth cell (175 cm or above and second column):Expected = (36 * 28) / 80 = 1008 / 80 = 12.6So all expected counts are above 5, which meets the assumption for the chi-squared test. So the test is valid.Therefore, since the calculated chi-squared statistic is approximately 9.67, which is greater than the critical value of 7.879 at the 0.005 significance level, we reject the null hypothesis. So yes, we can be 99.5% confident that there is an association between height and lung capacity levels.Now moving on to question (2). A fitness test group consists of 6 male students: 1 unqualified, 4 good, and 1 excellent. The scoring is: unqualified - 0, qualified - 1, good - 2, excellent - 3. Two students are selected randomly. Let X be the sum of their scores. Find the probability distribution and the mean of X.First, let's clarify the composition. There are 6 students: 1 unqualified (0 points), 4 good (2 points each), and 1 excellent (3 points). So the scores are: 0, 2, 2, 2, 2, 3.We need to select two students at random and find the probability distribution of the sum X. Then compute the mean of X.First, let's list all possible pairs and their sums. Since the students are distinct, even though some have the same score, the pairs are combinations of two different students. However, since four students have a score of 2, they are indistinct except for their identity.But for probability purposes, since all pairs are equally likely, we can compute the number of pairs that result in each possible sum and divide by the total number of pairs.Total number of pairs is C(6, 2) = 15.Possible sums X can take:- 0 + 2 = 2- 0 + 3 = 3- 2 + 2 = 4- 2 + 3 = 5- 3 + 3 = 6 (But there's only one student with 3, so this is impossible)Wait, let's list all possible pairs:1. Unqualified (0) and Good (2): Sum = 22. Unqualified (0) and Excellent (3): Sum = 33. Good (2) and Good (2): Sum = 44. Good (2) and Excellent (3): Sum = 5That's all. Since there are no two excellent students, sum of 6 is not possible. So possible values for X: 2, 3, 4, 5.Now, we need to compute how many pairs result in each sum.Number of pairs with sum 2: These are pairs of the unqualified student (0) with any of the 4 good students (2). There are 4 such pairs.Sum = 3: Pair of unqualified (0) with excellent (3). There is 1 such pair.Sum = 4: Pairs of two good students (2 + 2). The number of ways to choose 2 good students out of 4: C(4, 2) = 6.Sum = 5: Pairs of a good student (2) with the excellent student (3). There are 4 good students, so 4 such pairs.Check total pairs: 4 + 1 + 6 + 4 = 15, which matches C(6,2) = 15. Good.Therefore, the probability distribution is:X | P(X)---|---2 | 4/153 | 1/154 | 6/155 | 4/15Simplify the fractions:2: 4/153: 1/154: 6/15 = 2/55: 4/15Alternatively, leave as is.Now, the mean (expected value) of X is calculated as E[X] = Σ [x * P(X=x)].Compute each term:For X=2: 2*(4/15) = 8/15For X=3: 3*(1/15) = 3/15 = 1/5For X=4: 4*(6/15) = 24/15 = 8/5For X=5: 5*(4/15) = 20/15 = 4/3Now sum them up:8/15 + 1/5 + 8/5 + 4/3Convert all to fifteenths:8/15 + 3/15 + 24/15 + 20/15 = (8 + 3 + 24 + 20)/15 = 55/15 = 11/3 ≈ 3.6667Alternatively, compute as decimals:8/15 ≈ 0.53331/5 = 0.28/5 = 1.64/3 ≈ 1.3333Adding them: 0.5333 + 0.2 = 0.7333; 1.6 + 1.3333 = 2.9333; total 0.7333 + 2.9333 = 3.6666..., which is 11/3.So the mean is 11/3.Alternatively, perhaps there's another way to compute the expected value without enumerating all possibilities. The expected value of the sum is the sum of the expected values of each individual score, but since we are selecting two without replacement, the covariance needs to be considered. However, for linearity of expectation, even with dependence, E[X + Y] = E[X] + E[Y]. So the expected value of the sum is 2 * E[individual score], where E[individual score] is the average score of one student.Wait, let's check that. Since we are picking two students, the expected sum is E[X + Y] = E[X] + E[Y], where X and Y are the scores of the two students. Since the selection is without replacement, but expectation is linear regardless of dependence. So the expected sum is 2 * (average score of one student).First, compute the average score of one student:There are 6 students: 1 with 0, 4 with 2, 1 with 3.Total scores: 0 + 4*2 + 1*3 = 0 + 8 + 3 = 11.Average score per student: 11/6 ≈ 1.8333.Therefore, expected sum is 2 * (11/6) = 22/6 = 11/3 ≈ 3.6667, which matches our previous result. So this is a good cross-check.Therefore, the mean is indeed 11/3.So putting it all together:Probability distribution:X | P(X)---|---2 | 4/153 | 1/154 | 2/55 | 4/15Mean: 11/3.Wait, but in the problem statement, it mentions "qualified - 1 point". Wait, the group consists of 1 unqualified, 4 good, and 1 excellent. So no students with "qualified" (1 point). So in the group, scores are 0, 2, 2, 2, 2, 3. So the possible sums are 0+2, 0+3, 2+2, 2+3. Which we have already accounted for. So that's correct.So yes, the probability distribution is as above, and the mean is 11/3.Final Answer(1) boxed{Yes}(2) The probability distribution of ( X ) is:[begin{array}{c|c}X & P(X) hline2 & dfrac{4}{15} 3 & dfrac{1}{15} 4 & dfrac{2}{5} 5 & dfrac{4}{15} end{array}]The mean of ( X ) is boxed{dfrac{11}{3}}.</think>To determine if there is an association between the height of male students and their lung capacity levels at a 99.5% confidence level, we perform a chi-squared test using the given contingency table:| Height | Lung Capacity Level | Total ||------------------|---------------------|-------|| Below 175 cm | 22 | 22 | 44 || 175 cm or above | 30 | 6 | 36 || Total | 52 | 28 | 80 |Using the formula for the chi-squared statistic ( K^2 ):[K^2 = frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}]where ( a = 22 ), ( b = 22 ), ( c = 30 ), ( d = 6 ), and ( n = 80 ):1. Calculate ( (ad - bc) ): [ ad - bc = (22 cdot 6) - (22 cdot 30) = 132 - 660 = -528 ]2. Square the result: [ (-528)^2 = 278784 ]3. Multiply by ( n ): [ 80 cdot 278784 = 22302720 ]4. Calculate the denominator: [ (a+b)(c+d)(a+c)(b+d) = 44 cdot 36 cdot 52 cdot 28 = 2306304 ]5. Compute ( K^2 ): [ K^2 = frac{22302720}{2306304} approx 9.67 ]The critical value for ( alpha = 0.005 ) (99.5% confidence) is 7.879. Since ( 9.67 > 7.879 ), we reject the null hypothesis, concluding there is a significant association.For the probability distribution and mean of ( X ) (sum of scores of two randomly selected students):- Scores: 0, 2, 2, 2, 2, 3- Possible sums: 2, 3, 4, 5- Total pairs: ( binom{6}{2} = 15 )Probability distribution:- ( X = 2 ): 4 pairs (0 and 2) → ( frac{4}{15} )- ( X = 3 ): 1 pair (0 and 3) → ( frac{1}{15} )- ( X = 4 ): 6 pairs (2 and 2) → ( frac{6}{15} = frac{2}{5} )- ( X = 5 ): 4 pairs (2 and 3) → ( frac{4}{15} )Mean calculation:[E[X] = 2 cdot frac{4}{15} + 3 cdot frac{1}{15} + 4 cdot frac{2}{5} + 5 cdot frac{4}{15} = frac{8}{15} + frac{3}{15} + frac{24}{15} + frac{20}{15} = frac{55}{15} = frac{11}{3}]Final answers:(1) boxed{Yes}(2) The probability distribution of ( X ):[begin{array}{c|c}X & P(X) hline2 & dfrac{4}{15} 3 & dfrac{1}{15} 4 & dfrac{2}{5} 5 & dfrac{4}{15} end{array}]The mean of ( X ) is boxed{dfrac{11}{3}}.]

🔑: Solution:# Part 1:Given the contingency table and the formula for {K^2}, we can calculate the value of {K^2} as follows:- Let a=22 (Below 175 cm, Excellent), b=22 (Below 175 cm, Good), c=30 (175 cm or above, Excellent), and d=6 (175 cm or above, Good).- Total number of students, n=80.Using the formula for {K^2}:[K^2 = frac{80 times (22 times 6 - 22 times 30)^2}{44 times 36 times 52 times 28} approx 9.67]Comparing K^2 to the critical value from the table for P(K^{2}geqslant k) at 0.005:[9.67 > 7.879]Since the calculated K^2 value is greater than the critical value for 99.5% confidence (7.879), we reject the null hypothesis H_{0}. Therefore, we can be 99.5% confident that there is an association between the height of male students and their lung capacity levels.boxed{text{Yes, we can be } 99.5% text{ confident there is an association.}}# Part 2:For the probability distribution and mean of X, the scoring system assigns points as follows: unqualified - 0 points, qualified - 1 point, good - 2 points, excellent - 3 points. With 1 unqualified, 4 good, and 1 excellent, the possible sums of scores for two randomly selected students are 2, 3, 4, 5.- P(X=2) is the probability of selecting one unqualified and one good, or two good students:[P(X=2) = frac{C_1^1C_4^1}{C_6^2} = frac{4}{15}]- P(X=3) is the probability of selecting one unqualified and one excellent student:[P(X=3) = frac{C_1^1C_1^1}{C_6^2} = frac{1}{15}]- P(X=4) is the probability of selecting two good students:[P(X=4) = frac{C_{4}^{2}}{C_{6}^{2}} = frac{6}{15}]- P(X=5) is the probability of selecting one good and one excellent student:[P(X=5) = frac{C_{4}^{1}C_{1}^{1}}{C_{6}^{2}} = frac{4}{15}]The probability distribution of X is:| X | 2 | 3 | 4 | 5 ||-----|-----|-----|-----|-----|| P(X) | frac{4}{15} | frac{1}{15} | frac{6}{15} | frac{4}{15} |To find the mean E(X):[E(X) = 2 times frac{4}{15} + 3 times frac{1}{15} + 4 times frac{6}{15} + 5 times frac{4}{15} = frac{11}{3}]boxed{E(X) = frac{11}{3}}

🧠:[<think>Okay, let's see. I have this problem here where I need to find the value of P. The expression for P is the square root of (1988 multiplied by 1989 multiplied by 1990 multiplied by 1991 plus 1) and then subtract 1989 squared. The options are numbers from 1987 to 1990. Hmm, this looks a bit complicated with all those numbers close to each other. Maybe there's a pattern or a trick to simplify the calculation.First, let me write down the problem again to make sure I understand it correctly:P = √(1988×1989×1990×1991 + 1) - 1989²I need to compute this and then choose the correct answer from the options. The numbers are all consecutive around 1989. The product of four consecutive numbers plus one under a square root. That part might be a perfect square, which would simplify things. Let me think about how to handle that product.When I see a product of consecutive numbers like this, sometimes it's helpful to let a variable represent the middle term or something similar. Let's see: 1988, 1989, 1990, 1991. These are four consecutive numbers. Let me try to pair them up. Maybe (1988×1991) and (1989×1990). Let's compute those:First, 1988×1991. Let me compute that. Hmm, 1988×1991. Alternatively, since 1988 is 1991 - 3, maybe use the difference of squares formula? Wait, no, that might not help. Alternatively, 1988×1991 = (1990 - 2)(1990 + 1) = Not sure if that helps. Wait, maybe not. Let's compute it directly:1988×1991:First, 1988×2000 = 3,976,000. Then subtract 1988×9 = 17,892. So 3,976,000 - 17,892 = 3,958,108.Similarly, 1989×1990. Let's compute that. 1989×1990. Again, 1989 is 2000 - 11, so 1989×1990 = (2000 - 11)(1990) = 2000×1990 - 11×1990 = 3,980,000 - 21,890 = 3,958,110.Wait a second! So 1988×1991 = 3,958,108 and 1989×1990 = 3,958,110. These two products are very close to each other, differing by 2. So if I let x = 1989×1990 - 1 = 3,958,110 - 1 = 3,958,109, then 1988×1991 = x - 1 and 1989×1990 = x + 1? Wait, let's check:If 1989×1990 = 3,958,110, then x = 3,958,109. Then 1988×1991 = 3,958,108, which is x -1. So the product 1988×1989×1990×1991 is equal to (x - 1)(x + 1) = x² - 1. Therefore, 1988×1989×1990×1991 + 1 = x² - 1 + 1 = x². So the square root of that is x. Therefore, √(1988×1989×1990×1991 + 1) = x = 3,958,109. Therefore, P = 3,958,109 - 1989².Now, compute 1989². Let's compute that. 1989 squared. Let's recall that (a - b)² = a² - 2ab + b². Alternatively, 2000 - 11 = 1989. So 1989² = (2000 - 11)² = 2000² - 2×2000×11 + 11² = 4,000,000 - 44,000 + 121 = 4,000,000 - 44,000 is 3,956,000, plus 121 is 3,956,121. Therefore, 1989² = 3,956,121.Therefore, P = 3,958,109 - 3,956,121. Let's compute that. 3,958,109 minus 3,956,121. Subtract 3,956,000 from both: 3,958,109 - 3,956,000 = 2,109. Then subtract the remaining 121: 2,109 - 121 = 1,988. So P = 1,988. Wait, 1988? That's one of the options, option B. Hmm, so the answer is 1988. Let me verify that again to make sure I didn't make a mistake.Let me check the first part again. The product 1988×1989×1990×1991. By pairing 1988×1991 and 1989×1990, I found those products to be 3,958,108 and 3,958,110. Then their product is (x - 1)(x + 1) where x is 3,958,109. Then the product is x² - 1, so adding 1 gives x². Then the square root is x. Then P = x - 1989². Then x is 3,958,109 and 1989² is 3,956,121, so the difference is 3,958,109 - 3,956,121 = 1,988. So yes, 1988. Therefore, the answer is B.But let me check if there's another way to approach this problem in case I made a mistake in the arithmetic, which is easy with such large numbers. Maybe there's a formula or identity that can be used here.I recall that for four consecutive numbers, say n, n+1, n+2, n+3, the product n(n+1)(n+2)(n+3) + 1 is a perfect square. In fact, it's equal to (n² + 3n + 1)². Let me verify that:Let’s compute n(n+1)(n+2)(n+3) + 1. First, multiply n(n+3) = n² + 3n, and (n+1)(n+2) = n² + 3n + 2. Then the product is (n² + 3n)(n² + 3n + 2) = (n² + 3n)^2 + 2(n² + 3n). Then adding 1:(n² + 3n)^2 + 2(n² + 3n) + 1 = (n² + 3n + 1)^2. Yes, that's correct. So the product of four consecutive numbers plus one is the square of (n² + 3n + 1). Therefore, in this problem, n is 1988. So the product 1988×1989×1990×1991 + 1 = (1988² + 3×1988 + 1)^2. Therefore, the square root is 1988² + 3×1988 + 1. Therefore, P = 1988² + 3×1988 + 1 - 1989².Let me compute that. First, compute 1988² + 3×1988 + 1. Let's write 1988 as x. Then the expression becomes x² + 3x + 1. Then subtract (x + 1)². So P = x² + 3x + 1 - (x + 1)^2. Let's expand (x + 1)^2: x² + 2x + 1. Therefore, P = x² + 3x + 1 - x² - 2x -1 = x. So P = x. But x is 1988. Therefore, P = 1988. Therefore, the answer is B.Wow, that's a much simpler way. So using the identity, we can see that P is equal to n, which is 1988. That confirms the earlier calculation. So regardless of the method, the answer is 1988. Therefore, the correct option is B.Final Answerboxed{B}</think>To find the value of ( P ), we start with the given expression:[P = sqrt{1988 cdot 1989 cdot 1990 cdot 1991 + 1} - 1989^2]We recognize that the product of four consecutive numbers plus one can be simplified using a known identity. For four consecutive numbers ( n, n+1, n+2, n+3 ), the product ( n(n+1)(n+2)(n+3) + 1 ) is a perfect square, specifically ( (n^2 + 3n + 1)^2 ).Here, ( n = 1988 ). Applying the identity:[1988 cdot 1989 cdot 1990 cdot 1991 + 1 = (1988^2 + 3 cdot 1988 + 1)^2]Taking the square root, we get:[sqrt{1988 cdot 1989 cdot 1990 cdot 1991 + 1} = 1988^2 + 3 cdot 1988 + 1]Thus, substituting back into the expression for ( P ):[P = (1988^2 + 3 cdot 1988 + 1) - 1989^2]Let ( x = 1988 ). Then:[P = (x^2 + 3x + 1) - (x + 1)^2]Expanding ( (x + 1)^2 ):[(x + 1)^2 = x^2 + 2x + 1]Substituting back:[P = x^2 + 3x + 1 - (x^2 + 2x + 1) = x^2 + 3x + 1 - x^2 - 2x - 1 = x]Since ( x = 1988 ), we find:[P = 1988]Thus, the value of ( P ) is (boxed{B}).,<think>Alright, let's tackle this problem. So, we need to find the value of ( P = sqrt{1988 cdot 1989 cdot 1990 cdot 1991 + 1} + (-1989^2) ). The answer choices are from 1987 to 1990. Hmm, okay. Let's break this down step by step.First, the expression inside the square root looks complicated: ( 1988 cdot 1989 cdot 1990 cdot 1991 + 1 ). That's a product of four consecutive numbers plus one. I remember there's a trick with products of consecutive numbers, maybe something related to squaring or a quadratic. Let me think. If you have four consecutive numbers, maybe you can pair them up?Wait, another approach: sometimes when you have a product of consecutive numbers, you can let a variable represent the middle term or something. Let's see. Let me try to simplify ( 1988 cdot 1989 cdot 1990 cdot 1991 ). Maybe if I group them as (1988 * 1991) and (1989 * 1990). Let me compute each pair:First pair: 1988 * 1991. Let's calculate that. Hmm, 1988 * 1991. Let's note that 1991 = 1988 + 3, so maybe use the formula (a)(a + 3) = a² + 3a. So here, a = 1988. Then 1988*1991 = 1988² + 3*1988.Similarly, the second pair: 1989 * 1990. Here, 1990 = 1989 + 1, so that's 1989*(1989 + 1) = 1989² + 1989.So now the product becomes (1988² + 3*1988)(1989² + 1989). Hmm, not sure if this helps. Maybe there's another way.Alternatively, maybe set n = 1989. Let's try that substitution because the answer choices are around 1989, and the middle term in those four numbers is 1989 and 1990. Let's see:If n = 1989, then the four numbers are (n - 1), n, (n + 1), (n + 2). So the product is (n - 1) * n * (n + 1) * (n + 2). Then we need to compute sqrt[(n - 1) * n * (n + 1) * (n + 2) + 1] - n². Wait, the original expression is sqrt[product] + (-1989²). So in terms of n, it would be sqrt[(n - 1)n(n + 1)(n + 2) + 1] - n². And we need to see if that simplifies to something.Let me try expanding (n - 1)n(n + 1)(n + 2). Let's first compute (n - 1)(n + 2) and n(n + 1):(n - 1)(n + 2) = n² + 2n - n - 2 = n² + n - 2n(n + 1) = n² + nSo the product becomes (n² + n - 2)(n² + n). Let me let m = n² + n. Then the product is (m - 2)m = m² - 2m.So substituting back, (n² + n)² - 2(n² + n). Therefore, the product is (n² + n)^2 - 2(n² + n).Therefore, the expression inside the square root is (n² + n)^2 - 2(n² + n) + 1. Let's write that as:[(n² + n)^2 - 2(n² + n) + 1]. That looks like a quadratic in terms of (n² + n). Let me check:Let x = n² + n. Then the expression becomes x² - 2x + 1 = (x - 1)^2. Oh! So that's a perfect square. Therefore, the expression inside the square root simplifies to (n² + n - 1)^2.Therefore, sqrt[(n² + n - 1)^2] = |n² + n - 1|. Since n is a positive integer (1989), n² + n - 1 is positive, so we can drop the absolute value. So sqrt[...] = n² + n - 1.Therefore, the original expression P = (n² + n - 1) + (-n²) = (n² + n -1) - n² = n -1.Since n = 1989, then P = 1989 -1 = 1988. So the answer should be B) 1988.Wait, let me verify this again step by step to make sure I didn't make a mistake.First, substitution: let n = 1989, so numbers are (n-1), n, (n+1), (n+2). Product: (n-1)n(n+1)(n+2). We then paired (n-1)(n+2) = n² + n - 2 and n(n+1) = n² + n. Then multiplied those two to get (n² + n - 2)(n² + n) = (n² + n)^2 - 2(n² + n). Then inside the sqrt, that product plus 1 is (n² + n)^2 - 2(n² + n) + 1 = (n² + n -1)^2. Then sqrt of that is n² + n -1, subtract n² gives n -1. Therefore, P = n -1. Since n =1989, P=1988. So answer B.But let me check with actual numbers to make sure.Let me compute the product 1988*1989*1990*1991. Let's see:Alternatively, compute (1988*1991)*(1989*1990). Let me compute each pair:1988*1991: Let's compute 1988*1991. Let's note that (2000 - 12)*(2000 - 9) = 2000² - 2000*(12 +9) + 12*9 = 4,000,000 - 42,000 + 108 = 3,958,108. Wait, 2000 -12 is 1988, and 2000 -9 is 1991? Wait, 2000 -9 is 1991? 2000 -9 is 1991? Wait, 2000 -9 is 1991? No, 2000 -9 is 1991? Wait, 2000 - 9 = 1991, yes. So 1988 is 2000 -12, and 1991 is 2000 -9. So the product is (2000 -12)(2000 -9) = 2000² - 21*2000 + 108 = 4,000,000 -42,000 +108 = 3,958,108. Wait, let me check:Wait, (a - b)(a - c) = a² -a(b +c) + bc. So here a=2000, b=12, c=9. So:(2000 -12)(2000 -9) = 2000² -2000*(12+9) +12*9 = 4,000,000 -2000*21 +108 = 4,000,000 -42,000 +108 = 3,958,108.Okay, so 1988*1991 = 3,958,108.Now 1989*1990: Let's compute that. 1989*1990. Similarly, (2000 -11)*(2000 -10) = 2000² -2000*(11 +10) +11*10 = 4,000,000 -2000*21 +110 = same as above? Wait, 2000 -11=1989, 2000-10=1990. So:(2000 -11)(2000 -10) = 2000² -21*2000 +110 = 4,000,000 -42,000 +110 = 3,958,110. Wait, but 1989*1990: 1989*1990 = (2000 -11)(2000 -10) = 2000² -21*2000 +110 = same as before. But 1989*1990 is 1989*(1989 +1) = 1989² +1989. Let's compute 1989²:1989²: Let's compute (2000 -11)² = 2000² -2*2000*11 +11² = 4,000,000 -44,000 +121 = 3,956,121. Then add 1989: 3,956,121 +1,989 = 3,958,110. So that matches. So 1989*1990=3,958,110.So the product 1988*1989*1990*1991 = 3,958,108 * 3,958,110. Let's compute this. Wait, that's a big number, but note that these two numbers are 3,958,108 and 3,958,110. These are two consecutive even numbers: 3,958,108 and 3,958,110. So they are two apart. Let me note that 3,958,108 = x, then 3,958,110 = x +2. So their product is x(x +2) = x² +2x. But maybe another approach. Let me see:Alternatively, let me compute (3,958,109 -1)(3,958,109 +1) = (3,958,109)^2 -1. Wait, but 3,958,108 *3,958,110 = (3,958,109 -1)(3,958,109 +1) = (3,958,109)^2 -1. Therefore, the product is (3,958,109)^2 -1. Therefore, the product plus 1 is (3,958,109)^2 -1 +1 = (3,958,109)^2. Therefore, sqrt of that is 3,958,109. Then the expression P = 3,958,109 -1989².But wait, we need to compute sqrt[product +1] -1989² = 3,958,109 -1989². Let's compute 1989² again. As before, we had 1989²=3,956,121. Therefore, 3,958,109 -3,956,121= 1,988. Therefore, P=1,988. Which is option B. So that matches the previous result.Therefore, the answer is 1988. So the correct option is B.I think this is solid. The key insight was recognizing that the product of the four consecutive numbers can be rewritten as a square minus one, so when you add one, it becomes a perfect square. Then the square root cancels out, leaving a number which, when subtracted by 1989 squared, gives 1988. Clever problem!Final Answerboxed{B}</think>Given ( P = sqrt{1988 cdot 1989 cdot 1990 cdot 1991 + 1} + (-1989^2) ), we need to find the value of ( P ).First, we recognize that the product inside the square root involves four consecutive numbers. Let ( n = 1989 ). Then the numbers can be written as ( (n-1) ), ( n ), ( (n+1) ), and ( (n+2) ). The product becomes:[(n-1) cdot n cdot (n+1) cdot (n+2)]We can pair the terms as follows:[(n-1)(n+2) quad text{and} quad n(n+1)]Calculating each pair:[(n-1)(n+2) = n^2 + n - 2][n(n+1) = n^2 + n]Let ( m = n^2 + n ). The product becomes:[(m - 2) cdot m = m^2 - 2m]Thus, the expression inside the square root is:[m^2 - 2m + 1 = (m - 1)^2]Substituting back ( m = n^2 + n ), we get:[(n^2 + n - 1)^2]Taking the square root:[sqrt{(n^2 + n - 1)^2} = n^2 + n - 1]Therefore, the expression for ( P ) becomes:[P = (n^2 + n - 1) - n^2 = n - 1]Substituting ( n = 1989 ):[P = 1989 - 1 = 1988]Thus, the value of ( P ) is (boxed{B}).]

🔑:[Solution] Let x=1989, then the expression under the radical can be written asbegin{array}{l}(x-1) x(x+1)(x+2)+1 =left(x^{2}+xright)left(x^{2}+x-2right)+1 = {left[left(x^{2}+x-1right)+1right]left[left(x^{2}+x-1right)-1right]+1 } =left(x^{2}+x-1right)^{2}-1^{2}+1 =left(x^{2}+x-1right)^{2}, therefore quad P=sqrt{left(1989^{2}+1989-1right)^{2}}+left(-1989^{2}right)=1988 .end{array}Therefore, the correct choice is (B).

🧠:[<think>Okay, so I have this problem with two parts. Let's start with part (1). They give me two vectors, m and n. The vector m is (√3 sin2x + 2, cosx) and vector n is (1, 2cosx). The function f(x) is the dot product of m and n. I need to find the maximum and minimum values of f(x) on the interval [0, π/4]. Alright, let's break this down.First, the dot product of two vectors (a, b) and (c, d) is just ac + bd. So applying that here, f(x) would be (√3 sin2x + 2)*1 + (cosx)*(2cosx). Let me write that out:f(x) = (√3 sin2x + 2) * 1 + cosx * 2cosx= √3 sin2x + 2 + 2cos²xNow I need to simplify this expression so I can analyze its maximum and minimum. Let me recall some trigonometric identities. First, sin2x is 2sinx cosx, but maybe that's not immediately helpful. Also, cos²x can be written as (1 + cos2x)/2. Let me try substituting that in.So, f(x) = √3 sin2x + 2 + 2*(1 + cos2x)/2Simplify the 2 and the denominator 2: they cancel out, so it becomes:√3 sin2x + 2 + (1 + cos2x)= √3 sin2x + 3 + cos2xHmm, so that's √3 sin2x + cos2x + 3. Maybe I can combine the sine and cosine terms into a single trigonometric function. The formula for a*sinθ + b*cosθ is R sin(θ + φ), where R = √(a² + b²) and tanφ = b/a. Wait, or is it cosφ = a/R and sinφ = b/R? Let me check.Yes, if I have a*sinθ + b*cosθ, that can be written as R sin(θ + φ), where R is the amplitude. Alternatively, sometimes people write it as R cos(θ - φ). Let me confirm.Actually, let me use the identity: a sinθ + b cosθ = √(a² + b²) sin(θ + φ), where φ = arctan(b/a). Wait, maybe it's arctan(b/a) or arctan(a/b)? Let me recall.Suppose we have a sinθ + b cosθ = R sin(θ + φ). Expanding the right side: R sinθ cosφ + R cosθ sinφ. Comparing coefficients:a = R cosφb = R sinφTherefore, tanφ = b/a, so φ = arctan(b/a). Then R² = a² + b², so R = √(a² + b²).Alternatively, if we write it as R cos(θ - φ), then expanding:R cosθ cosφ + R sinθ sinφ, so coefficients would be a = R cosφ, b = R sinφ, same as before. So either way, the phase shift is determined by arctan(b/a) or arctan(a/b) depending on the form. Let's stick with the sine form here.In our case, the expression is √3 sin2x + cos2x. Let me see:Let me call θ = 2x. Then we have √3 sinθ + cosθ. So here, a = √3, b = 1. So R = √( (√3)^2 + 1^2 ) = √(3 + 1) = √4 = 2. Then φ = arctan(b/a) = arctan(1/√3) = π/6. Therefore, √3 sinθ + cosθ = 2 sin(θ + π/6). Substituting back θ = 2x:= 2 sin(2x + π/6)Therefore, f(x) = 2 sin(2x + π/6) + 3.So now f(x) is simplified to 2 sin(2x + π/6) + 3. Now, to find the maximum and minimum values of this function on the interval [0, π/4]. Since sine function has a range of [-1, 1], multiplying by 2 gives [-2, 2], then adding 3 gives [1, 5]. But we need to consider the interval [0, π/4] because the maximum and minimum could be at the endpoints or at critical points inside the interval.So first, let's find the derivative to check for critical points.f'(x) = derivative of 2 sin(2x + π/6) + 3 = 2*cos(2x + π/6)*2 = 4 cos(2x + π/6)Set derivative equal to zero to find critical points:4 cos(2x + π/6) = 0cos(2x + π/6) = 0Solutions are 2x + π/6 = π/2 + kπ, where k is integer.Solving for x:2x = π/2 - π/6 + kπ2x = (3π/6 - π/6) + kπ = (2π/6) + kπ = π/3 + kπx = π/6 + kπ/2Now, x must be in [0, π/4]. Let's check possible k values.For k = 0: x = π/6 ≈ 0.523, which is greater than π/4 ≈ 0.785. Wait, π/4 is approximately 0.785, but π/6 is approximately 0.523. Wait, 0.523 is less than 0.785? Wait, π is approximately 3.1416, so π/6 is about 0.523, and π/4 is about 0.785. So π/6 is within [0, π/4]?Wait, 0.523 is less than 0.785, so x = π/6 is within the interval [0, π/4]. Wait, no! Wait, if x = π/6 ≈ 0.523, and π/4 is approximately 0.785. So 0.523 is between 0 and 0.785. So yes, x = π/6 is in [0, π/4]. Wait, but π/6 is 30 degrees, and π/4 is 45 degrees. So π/6 is 30 degrees, which is less than 45 degrees, so yes, π/6 is in [0, π/4]. Wait, is π/6 equal to 30 degrees? Yes. π is 180 degrees, so π/6 is 30, π/4 is 45. So yes, π/6 is within [0, π/4].Therefore, x = π/6 is a critical point in the interval.For k = -1: x = π/6 - π/2 = -π/3, which is negative, so not in the interval.For k = 1: x = π/6 + π/2 = 2π/3 ≈ 2.094, which is beyond π/4.So the only critical point in [0, π/4] is x = π/6.Therefore, we need to evaluate f(x) at x = 0, x = π/6, and x = π/4.Compute f(0):f(0) = 2 sin(0 + π/6) + 3 = 2 sin(π/6) + 3 = 2*(1/2) + 3 = 1 + 3 = 4.f(π/6):First, compute 2x + π/6 when x = π/6: 2*(π/6) + π/6 = π/3 + π/6 = π/2.So f(π/6) = 2 sin(π/2) + 3 = 2*1 + 3 = 5.f(π/4):Compute 2x + π/6 when x = π/4: 2*(π/4) + π/6 = π/2 + π/6 = (3π/6 + π/6) = 4π/6 = 2π/3.So f(π/4) = 2 sin(2π/3) + 3. Sin(2π/3) is sin(π - π/3) = sin(π/3) = √3/2. Therefore:f(π/4) = 2*(√3/2) + 3 = √3 + 3 ≈ 1.732 + 3 ≈ 4.732.But let's compute exactly. So √3 ≈ 1.732, so √3 + 3 ≈ 4.732, which is less than 5. Therefore, comparing the three values:f(0) = 4,f(π/6) = 5,f(π/4) ≈ 4.732.Therefore, the maximum value on [0, π/4] is 5 at x = π/6, and the minimum value is 4 at x = 0. Wait, but is 4 the minimum? Because f(π/4) is approximately 4.732, which is higher than 4. So the function starts at 4 when x = 0, goes up to 5 at x = π/6, then decreases slightly to about 4.732 at x = π/4. Therefore, the minimum is at x = 0, which is 4, and the maximum is 5 at x = π/6. So the answer for part (1) is maximum 5 and minimum 4.Wait, but let me confirm f(π/4):2 sin(2π/3) + 3. 2π/3 is 120 degrees, sin(120°) is √3/2, so 2*(√3/2) +3 = √3 +3 ≈ 1.732 + 3 = 4.732. Yes, so 4.732 is less than 5. So yes, maximum at π/6 is 5, and the minimum is at x =0, which is 4.Therefore, part (1) answer: maximum 5, minimum 4.Now part (2). In triangle ABC, sides a, b, c opposite angles A, B, C. Given f(A) =4, b=1, area is √3/2. Find a.First, recall f(x) = √3 sin2x + 2 + 2cos²x. Wait, no. Wait, in part (1), we found that f(x) can be written as 2 sin(2x + π/6) +3, but also as √3 sin2x + cos2x +3. However, in part (2), they mention f(A) =4, which is the same function. So f(A) =4. Let me check.Given f(A) =4. So 2 sin(2A + π/6) +3 =4. Therefore, 2 sin(2A + π/6) =1 => sin(2A + π/6)=1/2.Alternatively, if using the other form: √3 sin2A + cos2A +3 =4 => √3 sin2A + cos2A =1. Maybe either way.But perhaps using the first form is simpler. Let's see.From f(A) =4, which is 2 sin(2A + π/6) +3 =4. Then 2 sin(2A + π/6) =1 => sin(2A + π/6)=1/2.So 2A + π/6 = π/6 + 2πk or 5π/6 + 2πk.Solving for A:Case 1: 2A + π/6 = π/6 + 2πk => 2A = 2πk => A = πk.But since A is an angle in a triangle, 0 < A < π, so k=0 gives A=0, which is invalid. k=1 gives A=π, which is also invalid.Case 2: 2A + π/6 = 5π/6 + 2πk => 2A = 5π/6 - π/6 + 2πk = 4π/6 + 2πk = 2π/3 + 2πk => A = π/3 + πk.Again, 0 < A < π, so possible solutions:k=0: A = π/3.k=1: A = π/3 + π = 4π/3, which is more than π, invalid.Therefore, the only solution is A= π/3.So angle A is 60 degrees or π/3 radians.Given that, and side b=1, which is opposite angle B. The area is √3/2. We need to find side a, opposite angle A.In triangle ABC, area = (1/2)*b*c*sin A. Wait, area = (1/2)*ab sin C, but depending on which sides and angles we use. Let's recall the formula.Area = (1/2)*a*b*sin C, but here, if we know angle A, and sides adjacent to it? Wait, better to use the formula:Area = (1/2)*b*c*sin A.Wait, no, if sides b and c are adjacent to angle A, then the area would be (1/2)*b*c*sin A. But in triangle notation, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. So if we want to use sides adjacent to angle A, they would be sides b and c, but angle A is between sides b and c. Therefore, area is indeed (1/2)*b*c*sin A.Given that, the area is √3/2, so:(1/2)*b*c*sin A = √3/2We know b=1, A= π/3, so substituting:(1/2)*1*c*sin(π/3) = √3/2Compute sin(π/3) = √3/2.So:(1/2)*1*c*(√3/2) = √3/2Multiply left side:(√3/4)*c = √3/2Multiply both sides by 4/√3:c = (√3/2)*(4/√3) = 2So c=2.Now, we need to find side a. Since we know sides b and c, and angle A, we can use the Law of Cosines.Law of Cosines: a² = b² + c² - 2bc cos ASubstituting known values:a² = 1² + 2² - 2*1*2*cos(π/3)Compute cos(π/3) = 1/2.So:a² = 1 + 4 - 4*(1/2) = 5 - 2 = 3Therefore, a = √3Wait, let me check that again.Law of Cosines:a² = b² + c² - 2bc cos A= 1 + 4 - 2*1*2*(1/2)= 5 - 2*(1/2)*4Wait, 2*1*2 is 4, multiplied by cos A (1/2): 4*(1/2) =2Therefore, 1 + 4 =5, minus 2 gives 3. So a²=3, so a=√3.Alternatively, maybe use the Law of Sines.Law of Sines: a/sin A = b/sin B = c/sin CWe know a/sin(π/3) = b/sin B = 1/sin B = c/sin C = 2/sin CBut maybe that's more complicated. Since we already found a via Law of Cosines, and it gives a=√3.But let's verify with Law of Sines.From area formula, we found c=2. So sides b=1, c=2, angle A= π/3. So we can use Law of Cosines as above. Alternatively, using Law of Sines:a/sin A = b/sin BBut we don't know angle B. Alternatively, since we have sides b=1, c=2, angle A=60 degrees, area is √3/2, which is consistent with the area formula.Alternatively, if we use Heron's formula, but we need all sides. But since we found a=√3, let's check the area with sides a=√3, b=1, c=2.Wait, Heron's formula: perimeter p = (a + b + c)/2 = (√3 +1 +2)/2 = (3 +√3)/2.Area = √[p(p - a)(p - b)(p - c)]But this might be complicated. Alternatively, use the area formula with sides b and c and angle A.Which is (1/2)*b*c*sin A = (1/2)*1*2*sin(60°) = (1)* (√3/2) = √3/2, which matches the given area. So that checks out. Therefore, a=√3 is correct.So the answer is √3.Final Answer(1) The maximum value is boxed{5} and the minimum value is boxed{4}.(2) The value of ( a ) is boxed{sqrt{3}}.</think>Given the vectors (overset{→}{m} = (sqrt{3} sin 2x + 2, cos x)) and (overrightarrow{n} = (1, 2 cos x)), the function (f(x) = overset{→}{m} cdot overrightarrow{n}) is calculated as follows:[f(x) = (sqrt{3} sin 2x + 2) cdot 1 + cos x cdot 2 cos x][= sqrt{3} sin 2x + 2 + 2 cos^2 x]Using the trigonometric identity (cos^2 x = frac{1 + cos 2x}{2}), we simplify:[f(x) = sqrt{3} sin 2x + 2 + 1 + cos 2x][= sqrt{3} sin 2x + cos 2x + 3]This can be rewritten using the amplitude-phase form:[sqrt{3} sin 2x + cos 2x = 2 sin left(2x + frac{pi}{6}right)]Thus,[f(x) = 2 sin left(2x + frac{pi}{6}right) + 3](1) To find the maximum and minimum values of (f(x)) on ([0, frac{pi}{4}]):- The critical points are found by setting the derivative (f'(x) = 4 cos left(2x + frac{pi}{6}right)) to zero: [ cos left(2x + frac{pi}{6}right) = 0 implies 2x + frac{pi}{6} = frac{pi}{2} + kpi ] Solving for (x), we find (x = frac{pi}{6}) within the interval ([0, frac{pi}{4}]).- Evaluating (f(x)) at critical points and endpoints: - (f(0) = 4) - (fleft(frac{pi}{6}right) = 5) - (fleft(frac{pi}{4}right) = sqrt{3} + 3 approx 4.732)The maximum value is (5) and the minimum value is (4).(2) In (triangle ABC), given (f(A) = 4), (b = 1), and the area (frac{sqrt{3}}{2}):- Solving (2 sin left(2A + frac{pi}{6}right) + 3 = 4): [ sin left(2A + frac{pi}{6}right) = frac{1}{2} implies 2A + frac{pi}{6} = frac{5pi}{6} implies A = frac{pi}{3} ]- Using the area formula (frac{1}{2} bc sin A = frac{sqrt{3}}{2}) with (b = 1) and (A = frac{pi}{3}): [ frac{1}{2} cdot 1 cdot c cdot frac{sqrt{3}}{2} = frac{sqrt{3}}{2} implies c = 2 ]- Applying the Law of Cosines: [ a^2 = b^2 + c^2 - 2bc cos A = 1^2 + 2^2 - 2 cdot 1 cdot 2 cdot frac{1}{2} = 3 implies a = sqrt{3} ]Final answers:(1) The maximum value is (boxed{5}) and the minimum value is (boxed{4}).(2) The value of (a) is (boxed{sqrt{3}}).,<think>Alright, so I have this problem here with vectors and a function f(x) defined as their dot product. Then there are two parts to the problem. Let's start with part (1). First, I need to recall how to compute the dot product of two vectors. The dot product of vectors (m₁, m₂) and (n₁, n₂) is m₁n₁ + m₂n₂. So, given m→ = (√3 sin2x + 2, cosx) and n→ = (1, 2cosx), then f(x) = m→ · n→ should be (√3 sin2x + 2)*1 + cosx*(2cosx). Let me write that out:f(x) = (√3 sin2x + 2) * 1 + cosx * 2cosx= √3 sin2x + 2 + 2cos²xOkay, so f(x) simplifies to √3 sin2x + 2 + 2cos²x. Now, maybe I can simplify this further using trigonometric identities. Let's see. The term 2cos²x can be written using the double angle identity: cos²x = (1 + cos2x)/2. So substituting that in:2cos²x = 2*(1 + cos2x)/2 = 1 + cos2xSo now f(x) becomes:√3 sin2x + 2 + 1 + cos2x= √3 sin2x + cos2x + 3So f(x) = √3 sin2x + cos2x + 3. Hmm, that seems more manageable. Maybe I can combine the sine and cosine terms into a single sinusoidal function. Let's try that. The general form is A sinθ + B cosθ = C sin(θ + φ) or C cos(θ + φ), where C = √(A² + B²) and φ is some phase shift. Let's see here:We have √3 sin2x + cos2x. Let me consider this as A sin2x + B cos2x, where A = √3 and B = 1. Then the amplitude C would be √( (√3)² + 1² ) = √(3 + 1) = √4 = 2. Then, we can write this as 2 sin(2x + φ). Let me check the phase shift. Alternatively, maybe write it as 2 cos(2x - φ). Let's recall the identity:A sinθ + B cosθ = C sin(θ + φ) where tanφ = B/A? Wait, no, I think it's:A sinθ + B cosθ = C sinθ cosφ + C cosθ sinφ = C sin(θ + φ) if A = C cosφ and B = C sinφ. So then, tanφ = B/A. Wait, let's do that properly. Let's set:√3 sin2x + 1 cos2x = C sin(2x + φ)Expanding the right-hand side:C sin2x cosφ + C cos2x sinφComparing coefficients:C cosφ = √3C sinφ = 1So then, (C sinφ)/(C cosφ) = tanφ = 1/√3 => φ = π/6.Since C² = (√3)² + 1² = 4 => C = 2. So yes, that works. Therefore:√3 sin2x + cos2x = 2 sin(2x + π/6)Therefore, f(x) = 2 sin(2x + π/6) + 3Alright, so f(x) simplifies to 2 sin(2x + π/6) + 3. Now, to find the maximum and minimum values of f(x) on the interval [0, π/4]. The sine function has a maximum of 1 and a minimum of -1, so 2 sin(...) would have a max of 2 and min of -2, so adding 3, the maximum would be 5 and minimum 1. However, we need to check if the argument 2x + π/6 reaches the angles where sine is 1 or -1 within the interval x ∈ [0, π/4].So first, let's find the range of 2x + π/6 when x ∈ [0, π/4]. When x = 0: 2*0 + π/6 = π/6 ≈ 0.523 radiansWhen x = π/4: 2*(π/4) + π/6 = π/2 + π/6 = (3π + π)/6 = 4π/6 = 2π/3 ≈ 2.094 radiansSo the argument of sine goes from π/6 to 2π/3. So, in this interval, what's the maximum and minimum of sin(θ)?Let's consider the sine curve between π/6 and 2π/3. The maximum value occurs at θ = π/2, where sin(π/2) = 1. The minimum value in this interval is the smaller of sin(π/6) and sin(2π/3). But sin(π/6) = 1/2 and sin(2π/3) = sin(π - π/3) = sin(π/3) = √3/2 ≈ 0.866. So the minimum in the interval is 1/2 and the maximum is 1. Therefore, the maximum value of sin(2x + π/6) is 1, and the minimum is 1/2. Therefore, the maximum of f(x) is 2*1 + 3 = 5, and the minimum is 2*(1/2) + 3 = 1 + 3 = 4. Wait, that contradicts my initial thought. Wait, no: the minimum of sin(θ) here is 1/2, so 2*(1/2) + 3 = 1 + 3 = 4. So the minimum is 4 and maximum is 5. But wait, hold on. The minimum of sin(θ) over [π/6, 2π/3] is 1/2? Let me confirm. The sine function increases from π/6 to π/2, then decreases from π/2 to 2π/3. At π/6, sin(π/6) = 1/2. At π/2, it's 1. At 2π/3, sin(2π/3) = √3/2 ≈ 0.866. So the minimum is at θ = π/6, which is 1/2, and maximum at π/2, which is 1. Therefore, yes, the minimum of sin(θ) over this interval is 1/2, and the maximum is 1.Therefore, f(x) has maximum 2*1 + 3 = 5 and minimum 2*(1/2) + 3 = 1 + 3 = 4. So on [0, π/4], the maximum value of f(x) is 5 and the minimum is 4.Wait, but let me check with calculus as well, just to be sure. Maybe there is a critical point inside [0, π/4]?Given f(x) = 2 sin(2x + π/6) + 3. The derivative is f’(x) = 2*cos(2x + π/6)*2 = 4 cos(2x + π/6). Setting derivative to zero:4 cos(2x + π/6) = 0 => cos(2x + π/6) = 0Solutions for 2x + π/6 = π/2 + kπ, k integer.Thus,2x + π/6 = π/2 + kπ=> 2x = π/2 - π/6 + kπ = (3π/6 - π/6) + kπ = (2π/6) + kπ = π/3 + kπ=> x = π/6 + (kπ)/2Now, check if x is in [0, π/4].For k=0: x = π/6 ≈ 0.523, which is greater than π/4 ≈ 0.785? Wait, π is approximately 3.1416, so π/4 ≈ 0.785 and π/6 ≈ 0.523. Wait, π/6 is approximately 0.523, which is less than π/4 (≈0.785). Wait, π/6 is about 0.523, which is less than π/4. Wait, π/4 is 0.25π ≈0.785, and π/6 is ≈0.523, which is indeed less. Wait, but wait, 0.523 is less than 0.785, so x=π/6 is within [0, π/4]? Wait, no, π/6 ≈0.523, which is less than π/4≈0.785, so yes, x=π/6 is within the interval [0, π/4]. Wait, but hold on. Wait, no. Wait, π/4 is approximately 0.785, and π/6 is approximately 0.523, which is still less than π/4. So x=π/6 is inside the interval [0, π/4]. Therefore, there is a critical point at x=π/6. But wait, let me compute π/4. Since π≈3.1416, so π/4≈0.7854, and π/6≈0.5236. So yes, 0.5236 is less than 0.7854, so x=π/6 is inside [0, π/4]. Therefore, we have a critical point at x=π/6. Therefore, to find the extrema, we need to evaluate f(x) at x=0, x=π/6, and x=π/4.So compute f(0):f(0) = 2 sin(π/6) + 3 = 2*(1/2) + 3 = 1 + 3 = 4f(π/6):First, compute 2x + π/6 when x=π/6:2*(π/6) + π/6 = (2π/6 + π/6) = 3π/6 = π/2So f(π/6) = 2 sin(π/2) + 3 = 2*1 + 3 = 5f(π/4):2*(π/4) + π/6 = π/2 + π/6 = (3π + π)/6 = 4π/6 = 2π/3sin(2π/3) = sin(π - π/3) = sin(π/3) = √3/2 ≈0.866Thus, f(π/4) = 2*(√3/2) + 3 = √3 + 3 ≈1.732 + 3 ≈4.732But since √3 is approximately 1.732, so √3 + 3 ≈4.732. But this is less than 5. So comparing the values:At x=0: 4At x=π/6:5At x=π/4:≈4.732Therefore, the maximum is 5 at x=π/6, and the minimum is 4 at x=0. Wait, but earlier analysis suggested the minimum of sin(theta) was 1/2, leading to f(x) minimum 4, but here the minimum is at x=0. Hmm, but according to the calculus approach, the critical point at x=π/6 gives maximum 5, and endpoints x=0 and x=π/4 give 4 and ≈4.732. So the minimum is indeed 4 at x=0. Therefore, the answer for part (1) is maximum 5 and minimum 4.So part (1) done. Now part (2). In triangle ABC, sides a, b, c opposite angles A, B, C respectively. Given f(A)=4, b=1, and area of triangle is √3/2, find the value of a.First, recall f(x) is 2 sin(2x + π/6) + 3. So f(A)=4 means:2 sin(2A + π/6) + 3 = 4Subtract 3:2 sin(2A + π/6) = 1Divide by 2:sin(2A + π/6) = 1/2So, sin(theta) =1/2. The solutions are theta = π/6 + 2kπ or 5π/6 + 2kπ, for integer k.Therefore:2A + π/6 = π/6 + 2kπ => 2A = 2kπ => A = kπor2A + π/6 = 5π/6 + 2kπ => 2A = 4π/6 + 2kπ => 2A = 2π/3 + 2kπ => A = π/3 + kπBut since A is an angle in a triangle, it must be between 0 and π. So possible solutions:From first case: A = 0 or π, which are invalid in a triangle.From second case: A = π/3 + kπ. For k=0: A=π/3; for k=1: A=4π/3 which is more than π, invalid. So only A=π/3.Therefore, angle A is π/3 radians, or 60 degrees.Given that, and given side b=1 (opposite angle B), and area is √3/2. Need to find side a (opposite angle A=π/3).First, recall area of triangle is (1/2)*b*c*sin A. Wait, or (1/2)*a*b*sin C? Wait, area can be calculated in different ways depending on the sides and angles. Since we know angle A, and sides opposite to A and B, perhaps we can use the formula involving two sides and the included angle. But wait, we know angle A, and side b opposite angle B. Hmm, maybe use the Law of Sines?Law of Sines says a/sin A = b/sin B = c/sin C.We know a is the side opposite angle A=π/3, and b=1 is opposite angle B.So, a / sin(π/3) = 1 / sin B => a = sin(π/3) / sin B.But we need another equation to relate B and the area. The area is given as √3/2. Let's recall that area can also be expressed as (1/2)*a*b*sin C, where C is the angle between sides a and b. Wait, but maybe more useful here is using the formula involving sides b and c and angle A. Since we know angle A, and if we can express c in terms of other variables.Alternatively, area is (1/2)*b*c*sin A. Wait, angle A is between sides b and c? Not necessarily. Wait, the area can be written as (1/2)*ab sin C, where C is the angle between sides a and b. But maybe not. Let's clarify.Actually, area = (1/2)*b*c*sin A if angle A is between sides b and c. But in triangle ABC, angle A is opposite side a. So sides b and c are adjacent to angle A. So actually, yes, if we consider angle A, then sides b and c are adjacent to angle A, and the angle between sides b and c would be angle A. Wait, no. Wait, in triangle ABC, angle A is between sides b and c. Because in a triangle, each angle is between two sides. The side opposite angle A is a, so angle A is between sides b and c. Therefore, the area can indeed be expressed as (1/2)*b*c*sin A.So area = (1/2)*b*c*sin A = √3/2Given that b=1, angle A=π/3, so sin A = sin(π/3) = √3/2.So substitute into area formula:(1/2)*1*c*(√3/2) = √3/2Simplify left side:(1/2)*(√3/2)*c = √3/2Multiply terms:(√3/4)*c = √3/2Multiply both sides by 4/√3:c = (√3/2)*(4/√3) = (4/2) = 2Therefore, side c=2.Now, we can use Law of Sines to find a.Law of Sines: a / sin A = b / sin B = c / sin CWe know a / sin(π/3) = 1 / sin B = 2 / sin CWe need to find a. If we can find sin B, then we can compute a.Alternatively, since we know sides b=1, c=2, and angle A=π/3, we can use Law of Cosines to find side a. Wait, Law of Cosines: a² = b² + c² - 2bc cos AWait, yes, because angle A is between sides b and c. So:a² = 1² + 2² - 2*1*2*cos(π/3)Compute:a² = 1 + 4 - 4*(1/2) = 5 - 2 = 3Therefore, a = √3.Wait, that seems straightforward. Let me check if this makes sense.Alternatively, since we know sides b=1, c=2, and angle A=π/3 between them, then using Law of Cosines directly gives a² = 1 + 4 - 2*1*2*(1/2) = 5 - 2 = 3, so a=√3.But let me cross-verify with Law of Sines. From Law of Sines:a / sin(π/3) = c / sin CWe know c=2 and a=√3.So √3 / (√3/2) = 2 / sin C => (√3) / (√3/2) = 2 / sin C => 2 = 2 / sin C => sin C = 1 => C=π/2.Therefore, angle C is π/2, right angle. Then angle B = π - π/3 - π/2 = π - 5π/6 = π/6. Then sin B = sin(π/6) = 1/2, so from Law of Sines, a / sin(π/3) = 1 / (1/2) => a = sin(π/3) * 2 = (√3/2)*2 = √3. So same result. Therefore, the value of a is √3.So part (2) answer is √3.But let me check if all steps are correct. Given area is √3/2, with b=1, angle A=60 degrees. Using area formula:Area = (1/2)*b*c*sin A = (1/2)*1*c*(√3/2) = (√3/4)*c = √3/2 => c=2. Then using Law of Cosines, a² = 1 + 4 - 4*(1/2)=3, so a=√3. Correct. Therefore, the answer is √3.Final Answer(1) The maximum value is boxed{5} and the minimum value is boxed{4}.(2) The value of ( a ) is boxed{sqrt{3}}.</think>Given the vectors (overset{→}{m} = (sqrt{3} sin 2x + 2, cos x)) and (overrightarrow{n} = (1, 2 cos x)), the function (f(x) = overset{→}{m} cdot overrightarrow{n}) is calculated as follows:[f(x) = (sqrt{3} sin 2x + 2) cdot 1 + cos x cdot 2 cos x = sqrt{3} sin 2x + 2 + 2 cos^2 x]Using the trigonometric identity (cos^2 x = frac{1 + cos 2x}{2}), we get:[2 cos^2 x = 1 + cos 2x]Thus, the function simplifies to:[f(x) = sqrt{3} sin 2x + cos 2x + 3]This can be rewritten using the amplitude-phase form:[sqrt{3} sin 2x + cos 2x = 2 sin left(2x + frac{pi}{6}right)]Therefore, the function becomes:[f(x) = 2 sin left(2x + frac{pi}{6}right) + 3]To find the maximum and minimum values on ([0, frac{pi}{4}]):- The range of (2x + frac{pi}{6}) is from (frac{pi}{6}) to (frac{2pi}{3}).- The maximum value of (sin theta) in this interval is 1 (at (theta = frac{pi}{2})), and the minimum value is (frac{1}{2}) (at (theta = frac{pi}{6})).Thus, the maximum value of (f(x)) is:[2 cdot 1 + 3 = 5]And the minimum value is:[2 cdot frac{1}{2} + 3 = 4]For part (2), given (f(A) = 4), we solve:[2 sin left(2A + frac{pi}{6}right) + 3 = 4 implies sin left(2A + frac{pi}{6}right) = frac{1}{2}]The valid solution for (A) in a triangle is (A = frac{pi}{3}). Using the area formula for (triangle ABC):[text{Area} = frac{1}{2} cdot b cdot c cdot sin A = frac{sqrt{3}}{2}]Given (b = 1) and (sin A = frac{sqrt{3}}{2}), we find (c = 2). Using the Law of Cosines:[a^2 = b^2 + c^2 - 2bc cos A = 1^2 + 2^2 - 2 cdot 1 cdot 2 cdot frac{1}{2} = 3 implies a = sqrt{3}] Final Answer(1) The maximum value is (boxed{5}) and the minimum value is (boxed{4}).(2) The value of (a) is (boxed{sqrt{3}}).]

🔑:(1) Since overset{→}{m}=( sqrt{3}sin 2x+2,cos x) and overrightarrow{n}=(1,2cos x),then f(x)=overset{→}{m}cdot overrightarrow{n}= sqrt{3}sin 2x+2+2cos ^{2}x= sqrt{3}sin 2x+cos 2x+3=2sin (2x+ frac{pi}{6})+3.Thus, f(x) is increasing on [0, frac{pi}{6} ] and decreasing on [ frac{pi}{6} , frac{pi}{4} ]. f(0)=4, f( frac{pi}{6})=5 , f( frac{pi}{4})=3+ sqrt{3} .Hence, f_{min}=4 and f_{max}=5.(2) From f(A)=4, we have 2sin (2A+ frac{pi}{6})+3=4, so sin (2A+ frac{pi}{6})= frac{1}{2}.Since A is an internal angle of triangle ABC, frac{pi}{6} < 2A+ frac{pi}{6} < frac{13pi}{6}, and so 2A+ frac{pi}{6}= frac{5pi}{6}, which gives A= frac{pi}{3}.Given that S_{triangle ABC}= frac{sqrt{3}}{2} and b=1, we have frac{1}{2}bcsin A= frac{sqrt{3}}{2}, which implies c=2.Therefore, a^{2}=b^{2}+c^{2}-2bccos A=4+1-2times 2times 1times frac{1}{2}=3, and so a=boxed{sqrt{3}}.

🧠:[<think>Okay, so I need to figure out these probabilities for the lottery draw. Let's start by understanding the problem step by step. First, there are two boxes: A and B. Box A has 4 red balls and 6 white balls, so a total of 10 balls. Box B has 5 red balls and 5 white balls, totaling 10 as well. In each lottery draw, they pick 2 balls from each box, making a total of 4 balls. The prizes depend on how many red balls are drawn among these 4. For the first question, we need the probability of winning the first prize, which is all 4 balls being red. That means we need to pick 2 red balls from box A and 2 red balls from box B. Let me recall how to compute probabilities for combinations. The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. For combinations, the number of ways to choose k items from n is denoted as C(n, k) or "n choose k."Starting with box A: There are 4 red balls and 6 white. We need to pick 2 red balls from box A. The number of ways to choose 2 red balls from 4 is C(4, 2). The total number of ways to pick any 2 balls from box A is C(10, 2). Similarly, for box B: 5 red and 5 white. We need 2 red balls here as well, so the number of favorable outcomes is C(5, 2), and total outcomes are C(10, 2). Therefore, the probability of getting 2 red from box A is C(4,2)/C(10,2), and for box B, it's C(5,2)/C(10,2). Since the draws from box A and box B are independent, we multiply these probabilities together to get the total probability for the first prize.Let me compute these values:First, C(4,2) is 6, and C(10,2) is 45. So the probability for box A is 6/45, which simplifies to 2/15.For box B, C(5,2) is 10, and C(10,2) is also 45. So the probability here is 10/45, which simplifies to 2/9.Multiplying the two probabilities: (2/15) * (2/9) = 4/135. So the probability of winning the first prize is 4/135. Let me double-check that. Yes, 2/15 times 2/9 is indeed 4/135. Hmm, 4 divided by 135 is approximately 0.0296, which is about 2.96%. That seems reasonable.Moving on to the second question: the probability of winning the second prize, which is exactly 3 red balls. So, in the 4 balls drawn, 3 are red and 1 is white. How can this happen? Since we are drawing 2 from each box, the red and white balls can be distributed between the boxes in different ways. Let's think about the possible cases.Case 1: 2 red from box A and 1 red from box B, and 1 white from box B.Case 2: 1 red from box A, 1 white from box A, and 2 red from box B.Wait, because in each box, we draw 2 balls. So the total red balls can be 3 if:From box A: 2 red, from box B: 1 red and 1 white.Or from box A: 1 red and 1 white, from box B: 2 red.These are the two scenarios that result in a total of 3 red balls. So we need to calculate the probabilities for both cases and sum them up since they are mutually exclusive.First, calculate the probability for 2 red from A and 1 red and 1 white from B.Probability of 2 red from A: as before, C(4,2)/C(10,2) = 6/45 = 2/15.Probability of 1 red and 1 white from B: the number of ways is C(5,1)*C(5,1) = 5*5=25. The total number of ways is C(10,2)=45. So the probability is 25/45 = 5/9.So the combined probability for this case is (2/15)*(5/9) = 10/135.Second case: 1 red and 1 white from A, and 2 red from B.Probability of 1 red and 1 white from A: C(4,1)*C(6,1) = 4*6=24. Total ways C(10,2)=45. So probability is 24/45 = 8/15.Probability of 2 red from B: as before, C(5,2)/C(10,2)=10/45=2/9.Multiply these probabilities: (8/15)*(2/9)=16/135.Now, sum the two cases: 10/135 + 16/135 = 26/135.So the probability of the second prize is 26/135. Let me check if that adds up. 10 +16 is 26, over 135. Simplified, 26 and 135 have no common factors, so it's 26/135 ≈ 0.1926, about 19.26%. Seems plausible.Third question: probability of winning any prize. That is, the probability that there is at least 1 red ball. Alternatively, 1 minus the probability of no red balls (which is the probability of not winning a prize).But maybe we can compute it by adding up all the probabilities from first to fourth prize. However, perhaps it's easier to compute 1 - probability(0 red balls). Let's see.If we calculate the probability of 0 red balls, meaning all 4 balls are white. That requires 2 white from A and 2 white from B.Probability of 2 white from A: C(6,2)/C(10,2) = 15/45 = 1/3.Probability of 2 white from B: C(5,2)/C(10,2) = 10/45 = 2/9.Multiply them: (1/3)*(2/9) = 2/27. So probability of no red balls is 2/27. Therefore, probability of winning any prize is 1 - 2/27 = 25/27 ≈ 0.9259 or 92.59%.Wait, but let me confirm this. Alternatively, we can compute the probability of getting 1, 2, 3, or 4 red balls and sum them up. Let's check if that gives the same result.First, first prize: 4 red: 4/135.Second prize: 3 red: 26/135.Third prize: 2 red. Let's compute that. Similarly, the third prize is if there are exactly 2 red balls. The cases for this would be:From box A: 2 red, box B: 0 red.From box A: 1 red, box B: 1 red.From box A: 0 red, box B: 2 red.So three cases.Case 1: 2 red from A, 0 red from B.Probability: [C(4,2)/C(10,2)] * [C(5,0)*C(5,2)/C(10,2)]. Wait, C(5,0)*C(5,2) is not correct. Wait, 0 red from B means 2 white. So C(5,0)*C(5,2) is C(5,0) for red and C(5,2) for white. Wait, no, for box B, to get 0 red, you need to choose 2 white. Since there are 5 white balls. So number of ways is C(5,2). So probability is [C(4,2)/C(10,2)] * [C(5,2)/C(10,2)].Which is (6/45)*(10/45) = (6*10)/(45*45) = 60/2025 = 4/135. Wait, but that's the same as the first prize? Wait no, first prize is 2 red from A and 2 red from B. This case is 2 red from A and 0 red (i.e., 2 white) from B. So that's different.Wait, okay, so the probability for this case is (C(4,2)/C(10,2)) * (C(5,2)/C(10,2)) = (6/45)*(10/45) = 60/2025 = 4/135. Hmm, 60 divided by 2025 is 4/135? Let's check. 2025 divided by 15 is 135, 60 divided by 15 is 4. Yes, so 4/135.Case 2: 1 red from A and 1 red from B. Wait, but in total, that would be 1+1=2 red. So how does this translate? So, from A: 1 red and 1 white. From B: 1 red and 1 white.So, probability is [C(4,1)*C(6,1)/C(10,2)] * [C(5,1)*C(5,1)/C(10,2)].Calculating this:From A: 4*6=24 ways, total 45, so 24/45.From B: 5*5=25 ways, total 45, so 25/45.Multiply these: (24/45)*(25/45) = 600/2025 = 40/135 (divided numerator and denominator by 15). So 40/135.Case 3: 0 red from A, 2 red from B.Probability: [C(4,0)*C(6,2)/C(10,2)] * [C(5,2)/C(10,2)].Which is (15/45) * (10/45) = (15*10)/(45*45) = 150/2025 = 10/135.So total probability for third prize is 4/135 + 40/135 + 10/135 = 54/135 = 6/15 = 2/5. Wait, 54/135 simplifies to 18/45, then 6/15, then 2/5. Hmm, 2/5 is 0.4.Fourth prize is 1 red ball. Let's check that. For fourth prize, exactly 1 red. So possible cases:From A: 1 red, 1 white; from B: 0 red, 2 white.Or from A: 0 red, 2 white; from B: 1 red, 1 white.So two cases.Case 1: 1 red from A and 0 red from B.Probability: [C(4,1)*C(6,1)/C(10,2)] * [C(5,0)*C(5,2)/C(10,2)].Which is (24/45)*(10/45) = 240/2025 = 16/135.Wait, C(5,0)*C(5,2) is 1*10=10. So yes, 10/45.So (24/45)*(10/45) = (24*10)/(45*45) = 240/2025 = 16/135.Case 2: 0 red from A and 1 red from B.Probability: [C(6,2)/C(10,2)] * [C(5,1)*C(5,1)/C(10,2)].Which is (15/45)*(25/45) = (15*25)/2025 = 375/2025 = 25/135.So total probability for fourth prize is 16/135 + 25/135 = 41/135.Now, let's check if all these probabilities sum up to 1 minus the probability of no prize.First prize: 4/135 ≈ 0.0296Second prize:26/135 ≈ 0.1926Third prize:54/135 = 0.4Fourth prize:41/135 ≈ 0.3037No prize:2/27 ≈ 0.07407Adding them up: 0.0296 + 0.1926 + 0.4 + 0.3037 + 0.07407 ≈ 1.0 (approximately). Let's check exact fractions:4 +26 +54 +41 + (2/27 converted to 10/135) = 4 +26 +54 +41 +10 = 135. So 135/135=1. Perfect, that checks out.Therefore, the probability of winning a prize is 1 - 2/27 = 25/27 ≈ 0.9259, which matches the earlier calculation.So for the third question, the answer is 25/27.But let me verify once again. If the probability of no prize is 2/27, then the rest is prizes. So 1 - 2/27 = 25/27. That's correct.Alternatively, adding all the prize probabilities:First prize:4/135Second prize:26/135Third prize:54/135Fourth prize:41/135Sum: 4 +26 +54 +41 = 125/135 = 25/27. Yep, same result.So that's good.Therefore, the answers are:1. First prize: 4/1352. Second prize:26/1353. Any prize:25/27Wait, but let me confirm once again the second prize calculation. When we have 3 red balls, there are two cases:Case 1: 2 red from A and 1 red from B. Wait, but if you take 2 red from A (2 red) and 1 red from B, but you also take 1 white from B. Wait, but in total, from B you have 2 balls: 1 red and 1 white. So total red is 2 +1=3.Similarly, the other case is 1 red from A and 2 red from B. But in that case, from A you have 1 red and 1 white, and from B 2 red, so total red is 1 +2=3. So the two cases are correctly considered. Calculations:Case 1: (C(4,2)/C(10,2)) * (C(5,1)*C(5,1)/C(10,2)) = (6/45)*(25/45) = (150)/2025 = 10/135Case 2: (C(4,1)*C(6,1)/C(10,2)) * (C(5,2)/C(10,2)) = (24/45)*(10/45) = (240)/2025 = 16/135Adding them: 10 +16=26. 26/135. So that's correct.Yes, so all the calculations seem consistent.Final Answer1. The probability of winning the first prize is boxed{dfrac{4}{135}}.2. The probability of winning the second prize is boxed{dfrac{26}{135}}.3. The probability of winning a prize is boxed{dfrac{25}{27}}.</think>To calculate the probabilities for the lottery draw, we consider the following:1. First Prize (All 4 balls are red): - Probability of drawing 2 red balls from box A: (frac{binom{4}{2}}{binom{10}{2}} = frac{6}{45} = frac{2}{15}) - Probability of drawing 2 red balls from box B: (frac{binom{5}{2}}{binom{10}{2}} = frac{10}{45} = frac{2}{9}) - Combined probability: (frac{2}{15} times frac{2}{9} = frac{4}{135})2. Second Prize (Exactly 3 red balls): - Case 1: 2 red from A and 1 red from B - Probability for 2 red from A: (frac{6}{45}) - Probability for 1 red and 1 white from B: (frac{binom{5}{1} binom{5}{1}}{binom{10}{2}} = frac{25}{45}) - Combined probability: (frac{6}{45} times frac{25}{45} = frac{10}{135}) - Case 2: 1 red from A and 2 red from B - Probability for 1 red and 1 white from A: (frac{binom{4}{1} binom{6}{1}}{binom{10}{2}} = frac{24}{45}) - Probability for 2 red from B: (frac{10}{45}) - Combined probability: (frac{24}{45} times frac{10}{45} = frac{16}{135}) - Total probability: (frac{10}{135} + frac{16}{135} = frac{26}{135})3. Probability of Winning Any Prize: - Probability of no red balls (all white): - Probability of 2 white from A: (frac{binom{6}{2}}{binom{10}{2}} = frac{15}{45} = frac{1}{3}) - Probability of 2 white from B: (frac{binom{5}{2}}{binom{10}{2}} = frac{10}{45} = frac{2}{9}) - Combined probability: (frac{1}{3} times frac{2}{9} = frac{2}{27}) - Probability of winning any prize: (1 - frac{2}{27} = frac{25}{27}) Final Answers1. The probability of winning the first prize is (boxed{dfrac{4}{135}}).2. The probability of winning the second prize is (boxed{dfrac{26}{135}}).3. The probability of winning a prize is (boxed{dfrac{25}{27}}).,<think>Okay, so I need to figure out these probabilities for the lottery draw at the shopping mall. Let's start with the first question: the probability of winning the first prize, which requires all four balls drawn to be red. Hmm, so we have two boxes, A and B. From each box, we're drawing two balls. Let me break this down step by step.First, Box A has 4 red balls and 6 white balls, making a total of 10 balls. Box B has 5 red balls and 5 white balls, so also 10 balls total. For the first prize, we need all four balls to be red. That means we need to draw 2 red balls from Box A and 2 red balls from Box B. Since these are separate boxes, the events are independent, right? So the probability should be the probability of getting 2 reds from Box A multiplied by the probability of getting 2 reds from Box B.Let me recall how to calculate combinations. The number of ways to choose 2 red balls from Box A is C(4,2), since there are 4 red balls. The total number of ways to draw any 2 balls from Box A is C(10,2). Similarly, for Box B, the number of ways to choose 2 red balls is C(5,2), and the total number of ways is C(10,2). Then I can multiply these two probabilities together because the draws from each box are independent.Calculating for Box A:C(4,2) = 4! / (2! * (4-2)!) = (4*3)/(2*1) = 6.C(10,2) = 10! / (2! * 8!) = (10*9)/2 = 45.So the probability for Box A is 6/45 = 2/15.For Box B:C(5,2) = 5! / (2! * 3!) = (5*4)/2 = 10.C(10,2) = 45 as before.So probability for Box B is 10/45 = 2/9.Then the combined probability for both boxes is (2/15) * (2/9) = 4/135. Let me check that again. 2/15 multiplied by 2/9 is (2*2)/(15*9) = 4/135. Hmm, 135 divided by 4 is 33.75, so 4/135 is approximately 0.0296, which is about 2.96%. That seems low, but considering we need four red balls, maybe that's right.Okay, moving on to the second question: the probability of winning the second prize, which requires exactly three red balls. So this would be scenarios where we have 3 red and 1 white. But since we are drawing two from each box, the possible combinations are:1. 2 red from Box A and 1 red and 1 white from Box B.OR2. 1 red and 1 white from Box A and 2 red from Box B.These are the two distinct cases that result in a total of 3 red balls. These two cases are mutually exclusive, so we can add their probabilities together.Let's compute each case.First case: 2 red from A, 1 red and 1 white from B.From Box A, the probability of 2 red is the same as before, 6/45 = 2/15.From Box B, we need 1 red and 1 white. The number of ways to choose 1 red and 1 white is C(5,1)*C(5,1) = 5*5 = 25. The total number of ways is still C(10,2) = 45. So the probability is 25/45 = 5/9.So the probability for this case is (2/15) * (5/9) = 10/135.Second case: 1 red and 1 white from A, and 2 red from B.From Box A, the number of ways to get 1 red and 1 white is C(4,1)*C(6,1) = 4*6 = 24. The total number of ways is C(10,2) = 45. So probability is 24/45 = 8/15.From Box B, the probability of 2 red is 10/45 = 2/9 as before.So the probability for this case is (8/15) * (2/9) = 16/135.Now add both cases: 10/135 + 16/135 = 26/135. Simplifying, that's 26/135. Let me check if that can be reduced. 26 and 135 share a common factor of... 13*2 and 27*5. Nope, no common factors. So 26/135 ≈ 0.1926, around 19.26%. That seems plausible for the second prize.Third question: the probability of winning any prize. So that would be 1 minus the probability of not winning a prize. Not winning a prize is when all four balls are white. So maybe it's easier to calculate the probability of not winning and subtract that from 1.Alternatively, we could calculate the probabilities for each prize (first to fourth) and sum them up. But since the user asks for the probability of winning a prize, which includes first, second, third, fourth. Let me check.The problem statement says: "if all are red, first prize; if only 3 red, second prize; only 2 red, third prize; only 1 red, fourth prize; no red, no prize." So winning a prize includes all cases except when there are 0 red balls. Therefore, the probability of winning a prize is 1 - P(0 red balls). Let me verify that.Yes, because the only non-prize is when all four balls are white. So P(prize) = 1 - P(0 red). Alternatively, if we calculate P(1 red) + P(2 red) + P(3 red) + P(4 red). But either way, let's compute it.But maybe computing P(0 red) is easier. So, to have 0 red balls, all four must be white. That means 2 white from Box A and 2 white from Box B.Calculating this:For Box A, number of white balls is 6. So the number of ways to choose 2 white is C(6,2). Total ways in Box A is C(10,2).Similarly, Box B has 5 white balls, so C(5,2) ways. Total ways in Box B is C(10,2).So probability for Box A: C(6,2)/C(10,2) = (15)/45 = 1/3.Probability for Box B: C(5,2)/C(10,2) = (10)/45 = 2/9.So the probability of 0 red is (1/3)*(2/9) = 2/27. Therefore, probability of winning a prize is 1 - 2/27 = 25/27 ≈ 0.9259 or 92.59%. Hmm, that seems high. Let me check again.Wait, but let's confirm with another method. Let's compute all the probabilities and see if they sum to 1.We already have P(4 red) = 4/135 ≈ 0.0296P(3 red) = 26/135 ≈ 0.1926Now let's compute P(2 red). That would be cases where:- 2 red from A and 0 red from B- 1 red from A and 1 red from B- 0 red from A and 2 red from BBut wait, no. Wait, total red balls can be 2 in different combinations. Since we draw two from each box, the possible red counts per box can be:Case 1: 2 red from A and 0 red from B.Case 2: 1 red from A and 1 red from B.Case 3: 0 red from A and 2 red from B.But in total, these would sum to 2 red, 2 red, and 2 red? Wait, no. Let me think again.Wait, if you have two boxes, each contributing 0, 1, or 2 red balls. So the total number of red balls is the sum from both boxes.To get a total of 2 red balls, the combinations are:- 2 red from A and 0 from B.- 1 red from A and 1 from B.- 0 from A and 2 from B.So these are the three cases for P(2 red). Let's compute each.First case: 2 red from A, 0 from B.Probability for 2 red in A: 2/15 as before.Probability for 0 red in B: That's choosing 2 white from B. Which is C(5,2)/C(10,2) = 10/45 = 2/9.So probability is (2/15)*(2/9) = 4/135.Second case: 1 red from A and 1 red from B.Probability for 1 red in A: C(4,1)*C(6,1)/C(10,2) = 24/45 = 8/15.Probability for 1 red in B: C(5,1)*C(5,1)/C(10,2) = 25/45 = 5/9.Multiply these: (8/15)*(5/9) = 40/135.Third case: 0 red from A and 2 red from B.Probability for 0 red in A: C(6,2)/C(10,2) = 15/45 = 1/3.Probability for 2 red in B: 10/45 = 2/9.Multiply these: (1/3)*(2/9) = 2/27 ≈ 0.0741.Convert 2/27 to 135 denominator: 2/27 = 10/135.Wait, 2/27 is equal to 10/135? Wait, 27*5=135, so 2*5=10, yes. So 2/27 = 10/135.So total P(2 red) = 4/135 + 40/135 + 10/135 = 54/135 = 6/15 = 2/5. Hmm, 54 divided by 135 is 0.4, which is 2/5. Okay.Then P(2 red) is 2/5, which is 0.4. Then let's compute P(1 red). That's total of 1 red ball in all four. Which can be:- 1 red from A and 0 from B.- 0 from A and 1 from B.But since we are drawing two from each box, getting 1 red total would require:From A: 1 red and 1 white; From B: 0 red and 2 white.ORFrom A: 0 red and 2 white; From B: 1 red and 1 white.So two cases.First case: 1 red in A and 0 red in B.Probability for A: 8/15 (as before).Probability for B: 2/9 (as in 0 red, which is same as 2 white, which was 2/9).Multiply: (8/15)*(2/9) = 16/135.Second case: 0 red in A and 1 red in B.Probability for A: 1/3 (0 red).Probability for B: 5/9 (1 red and 1 white).Multiply: (1/3)*(5/9) = 5/27 ≈ 0.1852. Convert to 135 denominator: 25/135.So total P(1 red) = 16/135 + 25/135 = 41/135 ≈ 0.3037.Then P(0 red) is 2/27 ≈ 0.0741, as before.So let's sum all probabilities:P(4 red) = 4/135 ≈ 0.0296P(3 red) = 26/135 ≈ 0.1926P(2 red) = 54/135 ≈ 0.4P(1 red) = 41/135 ≈ 0.3037P(0 red) = 10/135 ≈ 0.0741 (Wait, hold on, earlier we calculated P(0 red) as 2/27 ≈ 0.0741, which is 10/135. Wait, but 2/27 is 10/135. So 10/135 ≈ 0.0741)Wait, but in the case of P(2 red), I had 54/135, which is 0.4. Then adding up all these:0.0296 + 0.1926 + 0.4 + 0.3037 + 0.0741 ≈ 0.0296 + 0.1926 = 0.2222; 0.2222 + 0.4 = 0.6222; 0.6222 + 0.3037 = 0.9259; 0.9259 + 0.0741 = 1.0. Perfect, that sums to 1. So that checks out.Therefore, the probability of winning a prize is 1 - P(0 red) = 1 - 10/135 = 125/135 = 25/27 ≈ 0.9259, which is 92.59%.Wait, but 135 - 10 is 125, so 125/135 simplifies to 25/27, yes.Alternatively, adding up the probabilities of 4,3,2,1 red: 4 + 26 + 54 + 41 = 125. So 125/135 = 25/27. Correct.Therefore, the answer to part 3 is 25/27.But let me confirm once again for the third prize. Wait, in the problem statement, third prize is 2 red balls. So the probability is 54/135, which is 2/5. But according to our calculation for the third question, the probability of winning any prize is 25/27. But 25/27 is approximately 0.9259. So if we sum P(4) + P(3) + P(2) + P(1) = 4 + 26 + 54 + 41 = 125. 125/135 = 25/27. Correct.So all steps check out. The key steps were breaking down each case for the number of red balls, calculating the probabilities for each box, multiplying the independent probabilities, and then summing the appropriate cases. Then verifying that all probabilities add up to 1, which they do, so that gives confidence in the results.Final Answer1. The probability of winning the first prize is boxed{dfrac{4}{135}}.2. The probability of winning the second prize is boxed{dfrac{26}{135}}.3. The probability of winning a prize is boxed{dfrac{25}{27}}.</think>To calculate the probabilities for the lottery draw, we consider the following steps:1. Probability of winning the first prize (all 4 red balls): - From Box A (4 red, 6 white), the probability of drawing 2 red balls: [ frac{binom{4}{2}}{binom{10}{2}} = frac{6}{45} = frac{2}{15} ] - From Box B (5 red, 5 white), the probability of drawing 2 red balls: [ frac{binom{5}{2}}{binom{10}{2}} = frac{10}{45} = frac{2}{9} ] - Combined probability: [ left(frac{2}{15}right) times left(frac{2}{9}right) = frac{4}{135} ]2. Probability of winning the second prize (exactly 3 red balls): - Case 1: 2 red from Box A and 1 red from Box B - Probability from Box A: (frac{2}{15}) - Probability from Box B (1 red, 1 white): [ frac{binom{5}{1} binom{5}{1}}{binom{10}{2}} = frac{25}{45} = frac{5}{9} ] - Combined probability for Case 1: [ left(frac{2}{15}right) times left(frac{5}{9}right) = frac{10}{135} ] - Case 2: 1 red from Box A and 2 red from Box B - Probability from Box A (1 red, 1 white): [ frac{binom{4}{1} binom{6}{1}}{binom{10}{2}} = frac{24}{45} = frac{8}{15} ] - Probability from Box B: (frac{2}{9}) - Combined probability for Case 2: [ left(frac{8}{15}right) times left(frac{2}{9}right) = frac{16}{135} ] - Total probability: [ frac{10}{135} + frac{16}{135} = frac{26}{135} ]3. Probability of winning any prize: - Probability of not winning (0 red balls): - From Box A (2 white): [ frac{binom{6}{2}}{binom{10}{2}} = frac{15}{45} = frac{1}{3} ] - From Box B (2 white): [ frac{binom{5}{2}}{binom{10}{2}} = frac{10}{45} = frac{2}{9} ] - Combined probability: [ left(frac{1}{3}right) times left(frac{2}{9}right) = frac{2}{27} ] - Probability of winning a prize: [ 1 - frac{2}{27} = frac{25}{27} ] Final Answers1. The probability of winning the first prize is boxed{dfrac{4}{135}}.2. The probability of winning the second prize is boxed{dfrac{26}{135}}.3. The probability of winning a prize is boxed{dfrac{25}{27}}.]

🔑:Solution:1. The probability of a customer winning the first prize in one lottery draw:P_{1}= frac { C_{ 4 }^{ 2 } C_{ 5 }^{ 2 }}{ C_{ 10 }^{ 2 } C_{ 10 }^{ 2 }} = boxed{frac {4}{135}}. 2. The probability of a customer winning the second prize in one lottery draw:P_{2}= frac { C_{ 4 }^{ 2 } C_{ 5 }^{ 3 } C_{ 5 }^{ 1 }+ C_{ 4 }^{ 1 } C_{ 6 }^{ 1 } C_{ 5 }^{ 2 }}{ C_{ 10 }^{ 2 } C_{ 10 }^{ 2 }} = frac {78}{405} = boxed{frac {26}{135}}. 3. The probability of a customer winning a prize in one lottery draw:P_{3}= frac { C_{ 6 }^{ 2 } C_{ 5 }^{ 2 }}{ C_{ 10 }^{ 2 } C_{ 10 }^{ 2 }} = boxed{frac {75}{81}}.