🧠:[<think>Okay, let's see. The problem is to form the equation of a plane that intercepts segment OA = 3 on the Ox axis and is perpendicular to the vector N = {2, -3, 1}. Hmm, so I need to find the equation of a plane with these two conditions. Let me recall how to do this.First, I remember that the general equation of a plane is Ax + By + Cz + D = 0, where the vector N = {A, B, C} is the normal vector to the plane. In this case, the given normal vector is {2, -3, 1}, so the equation should be 2x - 3y + z + D = 0. But I need to find D such that the plane intercepts OA = 3 on the Ox axis. Wait, intercepting OA = 3 on the Ox axis means that the plane crosses the x-axis at the point (3, 0, 0), right? Because OA is along the Ox axis, so the intercept is at x=3, y=0, z=0. So if I plug that point into the plane equation, it should satisfy it. Let me check that.Plugging (3, 0, 0) into 2x - 3y + z + D = 0 gives 2*3 - 3*0 + 0 + D = 0. That simplifies to 6 + D = 0, so D = -6. Therefore, the equation of the plane is 2x - 3y + z - 6 = 0. Is that all? Let me confirm.Alternatively, maybe I can use the intercept form of the plane equation. The intercept form is x/a + y/b + z/c = 1, where a, b, c are the intercepts on the x, y, z axes respectively. But in this case, we only know the intercept on the x-axis, which is OA = 3, so a = 3. But we also know the normal vector, which might not align with the intercept form directly. Maybe combining both pieces of information would be better.Wait, since the plane is perpendicular to the normal vector N = {2, -3, 1}, the coefficients in the plane equation correspond to the normal vector. So we can start with 2x - 3y + z = D, and then use the intercept to find D. That seems straightforward. As before, substituting the point (3, 0, 0) gives 2*3 -3*0 +0 = D ⇒ D =6. Wait, but then the equation would be 2x -3y + z =6. Which would make the intercepts when y and z are zero, x = 6/2 = 3, which matches OA=3. So that works. But then, in the general equation, the standard form is Ax + By + Cz + D =0, so moving 6 to the left gives 2x -3y + z -6 =0, same as before. So that seems correct. Let me verify if this plane is indeed perpendicular to N. The normal vector of the plane 2x -3y + z -6 =0 is {2, -3, 1}, which matches the given N. So that's good. And the x-intercept is at x=3 when y=0 and z=0. So that's exactly OA=3. So the two conditions are satisfied. Therefore, the equation should be 2x -3y + z -6 =0. Alternatively, maybe I can derive the plane equation using the point-normal form. The point-normal form is N ⋅ (r - r₀) =0, where N is the normal vector, r is a general point on the plane, and r₀ is a specific point on the plane. Here, the specific point r₀ is (3,0,0), since that's where the plane intercepts the x-axis. So substituting into the formula:N ⋅ ( (x, y, z) - (3, 0, 0) ) =0Which becomes {2, -3, 1} ⋅ (x -3, y -0, z -0) =0Calculating the dot product:2*(x -3) + (-3)*(y) +1*(z) =0Expanding this:2x -6 -3y + z =0 ⇒ 2x -3y + z -6 =0Same result. So that confirms it. Therefore, the equation is 2x -3y + z -6 =0. Wait, just to ensure there's no mistake, let me check once more. The plane must pass through (3,0,0) and have normal vector {2, -3, 1}. So yes, substituting (3,0,0) into the equation gives 2*3 -3*0 +0 -6 =6 -6=0, which satisfies. Another point, for example, if x=0, y=0, then z=6. So the z-intercept is at (0,0,6). Similarly, if y=0, z=0, x=3. If x=0, z=0, then -3y -6=0 ⇒ y=-2, so y-intercept at (0, -2, 0). But the problem didn't mention those, just the x-intercept. So that's okay. The problem only required intercepting OA=3 on the Ox axis, which is satisfied, and being perpendicular to N, which is also satisfied. Therefore, the answer should be 2x -3y + z =6, or in standard form 2x -3y + z -6 =0. Depending on the preferred format. The problem says "form the equation", so either form is acceptable, but maybe they want it written as equal to 6. Let me check the problem statement again. It says "Form the equation of the plane...", so probably either form is okay, but maybe they prefer the form without the constant term on the right. Typically, the standard form is Ax + By + Cz + D =0. So writing 2x -3y + z -6 =0 is standard. Alternatively, some might write 2x -3y + z =6. Both are correct. The question didn't specify, so I can present either. But in the example given in the problem (Example 2), I should check if there's a standard answer. Wait, the user just said "Example 2." and then the problem. Maybe in the original context, but since it's not provided, maybe either form is okay. But let's stick with the standard form, which is 2x - 3y + z - 6 =0. So boxed as boxed{2x - 3y + z - 6 = 0} or boxed{2x - 3y + z = 6}. Either is acceptable, but since the problem mentions intercepting OA=3, maybe the intercept form is more related. However, the normal vector is given, so the standard form is more direct. Let me check another way. If using the intercept form, x/3 + y/b + z/c =1. But we also know the normal vector. The intercept form can be converted to standard form. The intercept form's normal vector would be coefficients when written as (1/3)x + (1/b)y + (1/c)z =1, so the normal vector would be (1/3, 1/b, 1/c). But the given normal vector is (2, -3, 1). Therefore, these should be scalar multiples. So (1/3, 1/b, 1/c) = k*(2, -3, 1). Therefore, 1/3 = 2k ⇒ k =1/6. Then 1/b = -3k = -3*(1/6) = -1/2 ⇒ b = -2. Similarly, 1/c =1*k =1/6 ⇒ c=6. Therefore, intercept form is x/3 + y/(-2) + z/6 =1, which can be rewritten as 2x -3y + z =6 (multiplying all terms by 6). So that's the same equation. So that's another way to get there. Therefore, either method leads to the same equation. So the answer is 2x -3y + z =6 or 2x -3y + z -6=0. Both are correct. Perhaps the problem expects the answer in the form with the constant on the right side. Let me check standard textbooks. Usually, the equation with the normal vector and a point is written as 2x -3y + z =6. But I think either form is acceptable. Since the user didn't specify, but the example given in the problem (Example 2) seems to be similar. Wait, the user wrote: "Example 2. Form the equation of the plane that intercepts segment OA=3 on the Ox axis and is perpendicular to the vector N = {2, -3, 1}." So in their example, maybe they expect the answer written as 2x -3y + z =6. Let me verify once again. If the plane intercepts the x-axis at (3,0,0), then substituting into 2x -3y + z =6 gives 2*3 =6, which is correct. The normal vector is {2, -3, 1}, which matches. So yes, this seems right. Therefore, the answer is boxed{2x - 3y + z = 6}.Final AnswerThe equation of the plane is boxed{2x - 3y + z = 6}.</think>To form the equation of the plane that intercepts segment (OA = 3) on the (Ox) axis and is perpendicular to the vector (vec{N} = {2, -3, 1}), we start with the general form of the plane equation using the normal vector. The normal vector gives us the coefficients (A), (B), and (C) in the plane equation (Ax + By + Cz + D = 0). Here, the normal vector is (vec{N} = {2, -3, 1}), so the equation is:[2x - 3y + z + D = 0]Next, we use the information that the plane intercepts the (Ox) axis at (OA = 3). This means the plane passes through the point ((3, 0, 0)). Substituting this point into the plane equation to find (D):[2(3) - 3(0) + 0 + D = 0][6 + D = 0][D = -6]Thus, the equation of the plane becomes:[2x - 3y + z - 6 = 0]We can also verify this using the point-normal form of the plane equation. The point-normal form is (vec{N} cdot (vec{r} - vec{r_0}) = 0), where (vec{r_0}) is a point on the plane. Using (vec{r_0} = (3, 0, 0)):[{2, -3, 1} cdot (x - 3, y - 0, z - 0) = 0][2(x - 3) - 3y + z = 0][2x - 6 - 3y + z = 0][2x - 3y + z - 6 = 0]Both methods lead to the same equation. Therefore, the equation of the plane is:[boxed{2x - 3y + z = 6}],<think>Okay, so I need to form the equation of a plane that intercepts a segment OA = 3 on the Ox axis and is perpendicular to the vector N = {2, -3, 1}. Hmm, let me recall how to do this. I think planes can be represented in different forms, like the general form or the intercept form. Also, since it's perpendicular to a given vector, that vector should be the normal vector of the plane. First, the normal vector is given as N = {2, -3, 1}. So the general equation of the plane should be 2x - 3y + z + D = 0, right? Because the coefficients of x, y, z in the plane equation correspond to the normal vector components. Now, I need to find D such that the plane intercepts the Ox axis at OA = 3. Wait, intercepting the Ox axis at OA = 3 means that the point A is (3, 0, 0) on the x-axis. So the plane passes through the point (3, 0, 0). If I substitute this point into the plane equation, I can solve for D. Let's do that. Plugging in x = 3, y = 0, z = 0 into 2x - 3y + z + D = 0:2*3 - 3*0 + 0 + D = 06 + D = 0So D = -6. Therefore, the equation of the plane should be 2x - 3y + z - 6 = 0. Let me check if this makes sense. The normal vector is correct. The plane passes through (3, 0, 0). Let me also check if the intercept on the x-axis is indeed 3. To find the x-intercept, set y = 0 and z = 0 in the equation:2x - 6 = 0 => x = 3. Yep, that's correct. But wait, is there another way to approach this? Maybe using the intercept form of the plane equation? The intercept form is x/a + y/b + z/c = 1, where a, b, c are the intercepts on the x, y, z axes respectively. But in this case, we only know the intercept on the x-axis, which is 3, so a = 3. But we don't know the intercepts on y and z axes. However, we do know the normal vector. Let me see if I can relate the intercept form to the normal vector. The intercept form can be converted to the general form. For example, x/3 + y/b + z/c = 1. Multiplying through by 3bc gives bc*x + 3c*y + 3b*z = 3bc. Then, comparing this to the general form 2x - 3y + z + D = 0, which is 2x - 3y + z = 6 (since D = -6). So, the coefficients of x, y, z in the intercept form multiplied out should be proportional to the normal vector components. Hmm, so if bc = 2, 3c = -3, and 3b = 1. Wait, maybe this is getting too complicated. Let's see. The normal vector of the intercept form x/a + y/b + z/c = 1 is actually the vector (1/a, 1/b, 1/c). Because the gradient (which is the normal vector) would be the coefficients of x, y, z. But in the intercept form, the equation is (x/a) + (y/b) + (z/c) = 1, so the coefficients are (1/a, 1/b, 1/c). Therefore, the normal vector is proportional to (1/a, 1/b, 1/c). But in our problem, the normal vector is given as (2, -3, 1). Therefore, (1/a, 1/b, 1/c) should be a scalar multiple of (2, -3, 1). Since a is 3, then 1/a = 1/3. But according to the normal vector, the x-component is 2. So 1/3 should be proportional to 2. That would mean that 1/3 = k*2, so k = 1/(6). Therefore, the normal vector from the intercept form would be (1/a, 1/b, 1/c) = (2k, -3k, k) = (2*(1/6), -3*(1/6), 1*(1/6)) = (1/3, -1/2, 1/6). But in that case, 1/a = 1/3, so a = 3, which matches. Then 1/b = -1/2, so b = -2, and 1/c = 1/6, so c = 6. Therefore, the intercept form would be x/3 + y/(-2) + z/6 = 1. Let's check if this is equivalent to the equation 2x - 3y + z - 6 = 0. Multiplying the intercept form by 6: 2x - 3y + z = 6. Yes, that's the same as 2x - 3y + z - 6 = 0. So both methods give the same result. Therefore, the equation of the plane is 2x - 3y + z - 6 = 0. Let me just verify once more. The normal vector is (2, -3, 1), which is correct. The plane passes through (3, 0, 0), which gives the x-intercept of 3, as required. If I set y and z to zero, x is 3. Similarly, if I wanted to find other intercepts, for example, y-intercept: set x=0, z=0, then -3y -6 =0 => y= -2, which matches b=-2 from before. And z-intercept: set x=0, y=0, then z -6 =0 => z=6, which matches c=6. So all intercepts make sense. Therefore, I think the answer is 2x - 3y + z - 6 = 0, which can also be written as 2x - 3y + z = 6. Either form is acceptable, but since the problem says "form the equation," either is fine. Probably the standard form with the constant term on the right is preferable, but both are correct. I don't see any mistakes in this reasoning. The key steps were recognizing the normal vector, using the intercept point to find the constant term, and verifying via intercept form. All steps align with the given conditions. Final AnswerThe equation of the plane is boxed{2x - 3y + z = 6}.</think>To form the equation of the plane that intercepts segment ( OA = 3 ) on the ( Ox ) axis and is perpendicular to the vector ( vec{N} = {2, -3, 1} ):1. Identify the normal vector: The given normal vector is ( vec{N} = {2, -3, 1} ). Therefore, the general form of the plane equation is ( 2x - 3y + z + D = 0 ).2. Determine the intercept point: The plane intercepts the ( Ox ) axis at ( OA = 3 ), meaning it passes through the point ( (3, 0, 0) ).3. Substitute the intercept point into the plane equation: [ 2(3) - 3(0) + 0 + D = 0 implies 6 + D = 0 implies D = -6 ]4. Form the equation: Substituting ( D = -6 ) into the general form gives: [ 2x - 3y + z - 6 = 0 ] This can be rewritten as: [ 2x - 3y + z = 6 ]5. Verification: Checking the intercepts: - ( x )-intercept: Set ( y = 0 ) and ( z = 0 ), gives ( 2x = 6 implies x = 3 ). - ( y )-intercept: Set ( x = 0 ) and ( z = 0 ), gives ( -3y = 6 implies y = -2 ). - ( z )-intercept: Set ( x = 0 ) and ( y = 0 ), gives ( z = 6 ).All intercepts are consistent with the given conditions. Therefore, the equation of the plane is:[boxed{2x - 3y + z = 6}]]

🔑:Solution. According to the condition, the point A(3,0,0) belongs to the desired plane. According to point 3^{circ}, the equation of this plane has the form2(x-3)-3(y-0)+(z-0)=0, quad text { or } quad 2 x-3 y+z-6=0

🧠:[<think>Okay, so I have this problem here where there are less than 100 students performing a group dance. The problem states that there are two different combinations of groups they can form. One combination is a group of 5 in the middle with the rest forming groups of 8 around the outer circle. The other combination is a group of 8 in the middle with the rest forming groups of 5 around the outer circle. The question is asking for the maximum number of students there can be, given that it's less than 100. So, we need to find the largest possible number under 100 that satisfies both these conditions.Let me start by breaking down the problem. Let's denote the total number of students as N, where N < 100.First combination: A group of 5 in the middle, and the remaining students (N - 5) form groups of 8 around the outer circle. So, N - 5 must be divisible by 8. That means (N - 5) mod 8 = 0. In other words, N ≡ 5 mod 8.Second combination: A group of 8 in the middle, and the remaining students (N - 8) form groups of 5 around the outer circle. So, N - 8 must be divisible by 5. That means (N - 8) mod 5 = 0. Therefore, N ≡ 8 mod 5. Simplifying 8 mod 5, we get 3, so N ≡ 3 mod 5.So, we need a number N < 100 such that:1. N ≡ 5 mod 82. N ≡ 3 mod 5And we need the largest such N under 100.So, this is a system of congruences. Let me write it down:N ≡ 5 (mod 8)N ≡ 3 (mod 5)We can solve this using the Chinese Remainder Theorem (CRT), but since the moduli here are 8 and 5, which are coprime (their GCD is 1), the CRT tells us there is a unique solution modulo 40 (since 8*5=40). Therefore, the solutions will repeat every 40 numbers. So, the possible solutions are N = x + 40k, where x is the smallest positive solution, and k is a non-negative integer.First, let's find the smallest N that satisfies both congruences.Let me solve the system:N ≡ 5 mod 8N ≡ 3 mod 5Express N as 8a + 5 for some integer a. Then substitute into the second equation:8a + 5 ≡ 3 mod 5Subtract 5 from both sides:8a ≡ -2 mod 5But -2 mod 5 is equivalent to 3 mod 5, so:8a ≡ 3 mod 5Since 8 ≡ 3 mod 5, this simplifies to:3a ≡ 3 mod 5Divide both sides by 3 (since 3 and 5 are coprime, division is allowed):a ≡ 1 mod 5Therefore, a = 5b + 1 for some integer b.Substituting back into N:N = 8*(5b + 1) + 5 = 40b + 8 + 5 = 40b + 13Therefore, the solutions are N ≡ 13 mod 40. So the numbers that satisfy both congruences are 13, 53, 93, 133, etc. But since N must be less than 100, the possible candidates are 13, 53, 93.But we need the maximum N under 100, which is 93. However, we need to verify if 93 indeed satisfies both original conditions.Let's check:First combination: Group of 5 in the middle, so remaining students = 93 - 5 = 88. 88 divided by 8 is 11, which is an integer. So that works.Second combination: Group of 8 in the middle, remaining students = 93 - 8 = 85. 85 divided by 5 is 17, which is also an integer. So that works as well.Hence, 93 is a valid solution. But wait, is there any higher number under 100? The next one would be 93 + 40 = 133, which is over 100, so 93 is indeed the largest possible.But let me double-check, maybe there are numbers between 53 and 93 that also satisfy both congruences? Wait, no. Because the solutions are spaced every 40. So starting at 13, then 53, then 93, then 133, etc. So between 53 and 93, there's a gap of 40. So 93 is indeed the next one after 53.But just to be thorough, maybe we can check another way. Let's list all numbers under 100 that are congruent to 5 mod 8:Starting from 5: 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, 101... but 101 is over 100, so stopping at 93.Similarly, numbers congruent to 3 mod 5 under 100: starting from 3: 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63, 68, 73, 78, 83, 88, 93, 98.Now, looking for common numbers in both lists. Let's see:First list (5 mod 8): 5,13,21,29,37,45,53,61,69,77,85,93Second list (3 mod 5): 3,8,13,18,23,28,33,38,43,48,53,58,63,68,73,78,83,88,93,98Intersection: 13,53,93. So these are the numbers common to both lists, so they satisfy both congruencies. Therefore, the largest one under 100 is 93.Therefore, the answer should be 93.But wait, just to be absolutely sure, let's confirm once again.For N=93:First case: 93 students. Group of 5 in the middle. Remaining 88. 88 divided by 8 is 11 groups. So yes.Second case: Group of 8 in the middle. Remaining 85. 85 divided by 5 is 17 groups. That works.So yes, 93 is correct.But just to check if 98 is possible. 98 is in the second list (3 mod 5), but is it in the first list? 98 mod 8: 98 divided by 8 is 12*8=96, remainder 2. So 98 ≡ 2 mod 8. Not 5 mod 8. So 98 is not in the first list. So 98 is invalid. So 93 is indeed the maximum.Therefore, the answer is 93.Final AnswerThe maximum number of students is boxed{93}.</think>Given that the number of students ( N ) is less than 100, we need to find the largest ( N ) such that:1. When forming a group of 5 in the middle, the remaining students form groups of 8. This implies ( N - 5 ) is divisible by 8, so ( N equiv 5 pmod{8} ).2. When forming a group of 8 in the middle, the remaining students form groups of 5. This implies ( N - 8 ) is divisible by 5, so ( N equiv 3 pmod{5} ).We solve the system of congruences:[begin{cases}N equiv 5 pmod{8} N equiv 3 pmod{5}end{cases}]Express ( N ) as ( 8a + 5 ). Substituting into the second congruence:[8a + 5 equiv 3 pmod{5} implies 8a equiv -2 pmod{5} implies 3a equiv 3 pmod{5} implies a equiv 1 pmod{5}]Thus, ( a = 5b + 1 ). Substituting back:[N = 8(5b + 1) + 5 = 40b + 13]The solutions are ( N equiv 13 pmod{40} ). The numbers under 100 are 13, 53, and 93. The largest is 93.Verification:- For ( N = 93 ): - Group of 5: ( 93 - 5 = 88 ), which is divisible by 8. - Group of 8: ( 93 - 8 = 85 ), which is divisible by 5.Thus, the maximum number of students is (boxed{93}).,<think>Okay, so I have this problem here where I need to find the maximum number of students less than 100 that can perform a group dance under two different combinations. The first combination is having a group of 5 in the middle and the rest forming groups of 8 around the outer circle. The second combination is having a group of 8 in the middle and the rest forming groups of 5 around the outer circle. I need to figure out the largest possible number of students that satisfies both these conditions. Let me try to break this down step by step.First, let's parse the problem again to make sure I understand it correctly. We have two scenarios:1. A central group of 5 students, with the remaining students arranged in groups of 8 in an outer circle.2. A central group of 8 students, with the remaining students arranged in groups of 5 in an outer circle.The total number of students in both cases must be the same, and this number is less than 100. We need to find the largest such number.So, mathematically, let me represent the total number of students as N, where N < 100.In the first case, there's a group of 5 in the middle, so the remaining students are N - 5. These remaining students must form groups of 8. So, N - 5 must be divisible by 8. In other words, N - 5 ≡ 0 mod 8. That gives us the equation:N ≡ 5 mod 8.Similarly, in the second case, there's a group of 8 in the middle, so the remaining students are N - 8. These must form groups of 5, meaning N - 8 must be divisible by 5. So:N ≡ 8 mod 5.But 8 mod 5 is equivalent to 3 mod 5, so:N ≡ 3 mod 5.Therefore, the problem reduces to finding the largest N < 100 such that:N ≡ 5 mod 8,andN ≡ 3 mod 5.So now, this is a system of two congruences. To solve this, I can use the Chinese Remainder Theorem (CRT). But before I jump into that, let me recall that CRT applies when the moduli are coprime. In this case, the moduli are 8 and 5. Since 8 and 5 are coprime (their greatest common divisor is 1), CRT tells us that there exists a unique solution modulo 40 (since 8*5=40). Therefore, the solutions will repeat every 40 numbers. So, the possible solutions are of the form N = 40k + m, where m is the smallest positive solution, and k is an integer.First, I need to find m, the smallest positive integer that satisfies both congruences.So, let's solve the system:N ≡ 5 mod 8,N ≡ 3 mod 5.Let me express N in terms of one of the congruences. Let's take N ≡ 5 mod 8. So, N can be written as N = 8a + 5, where a is a non-negative integer.Now, substitute this into the second congruence:8a + 5 ≡ 3 mod 5.Simplify:8a ≡ 3 - 5 mod 5,8a ≡ -2 mod 5,But -2 mod 5 is 3 mod 5, so:8a ≡ 3 mod 5.But 8 mod 5 is 3, so:3a ≡ 3 mod 5.Divide both sides by 3 (since 3 and 5 are coprime, division is allowed):a ≡ 1 mod 5.Therefore, a can be written as a = 5b + 1, where b is a non-negative integer.Substitute back into N:N = 8*(5b + 1) + 5 = 40b + 8 + 5 = 40b + 13.So, the solutions are N ≡ 13 mod 40. Therefore, the numbers satisfying both congruences are 13, 53, 93, 133, etc. But since we need N < 100, the possible numbers are 13, 53, and 93.Hence, the maximum N less than 100 is 93.Wait, but let me verify this because I need to make sure that both conditions hold for N=93.First condition: Group of 5 in the middle, rest in groups of 8.So, 93 - 5 = 88. 88 divided by 8 is 11. So, 11 groups of 8. That works.Second condition: Group of 8 in the middle, rest in groups of 5.93 - 8 = 85. 85 divided by 5 is 17. So, 17 groups of 5. That works too.So 93 satisfies both conditions and is less than 100. The next one would be 133, which is over 100, so 93 is the maximum.But wait, hold on. Let me check if there's a larger number less than 100. The solutions are 13, 53, 93. So 93 is the largest under 100. So yes, 93 is the answer.But let me think again—is there any mistake here? Let me check with 93.First arrangement: 5 in the center, so 93 - 5 = 88. 88 divided by 8 is 11. So, 11 outer groups. That's fine. 11*8=88. 88+5=93.Second arrangement: 8 in the center, so 93 - 8 = 85. 85 divided by 5 is 17. 17*5=85. 85+8=93. That also works. So 93 is correct.But let me check 53 as well to see the pattern.53 -5=48. 48/8=6. 6*8=48. 48+5=53.53 -8=45. 45/5=9. 9*5=45. 45+8=53. So that works too.Similarly, 13:13-5=8. 8/8=1. 1*8=8. 8+5=13.13-8=5. 5/5=1. 1*5=5. 5+8=13. Also works.So yes, the solutions are 13, 53, 93. So 93 is the largest under 100.But wait, the problem says "at most", so 93 is the answer.Wait, but just to be thorough, let's check if 93 is indeed less than 100. Yes, 93 is less than 100. So, the answer should be 93.But wait, hold on a second. Let me check if there's another possible number between 93 and 100 that might also satisfy the conditions. Since the modulus is 40, the next number after 93 would be 93 + 40 = 133, which is over 100. Therefore, 93 is indeed the largest one under 100.Alternatively, maybe I can approach this problem by considering the two conditions:N = 5 + 8a,andN = 8 + 5b,where a and b are positive integers. Then, equate them:5 + 8a = 8 + 5b.So, 8a - 5b = 3.We need to solve this Diophantine equation for integers a and b.Let me rearrange it:8a - 5b = 3.Let me try to find solutions for this equation.First, find particular solutions.We can write this as 8a = 5b + 3.Looking for integers a and b such that 5b + 3 is divisible by 8.Let me check for small values of b:b=1: 5*1 +3=8, which is divisible by 8. So, a=1.So, a=1, b=1 is a solution: N=5+8*1=13, which is the smallest solution.Then, the general solution can be found by adding the homogeneous solution.The homogeneous equation is 8a -5b=0. The solutions are a=5k, b=8k for integer k.Therefore, the general solution is:a = 1 + 5k,b = 1 + 8k,for integer k.Therefore, substituting back into N=5 +8a:N=5 +8*(1 +5k)=5 +8 +40k=13 +40k.Similarly, substituting into N=8 +5b:N=8 +5*(1 +8k)=8 +5 +40k=13 +40k.So, the solutions are N=13 +40k, where k is a non-negative integer. So N=13,53,93,133,...Since N must be less than 100, the maximum N is 93. Therefore, the answer is 93.Therefore, I can confirm that 93 is the correct answer. Thus, the maximum number of students is 93.Final AnswerThe maximum number of students is boxed{93}.</think>Given the problem of finding the maximum number of students less than 100 who can perform a group dance under two different combinations, we need to solve for the largest ( N ) such that:1. ( N - 5 ) is divisible by 8 (i.e., ( N equiv 5 pmod{8} )).2. ( N - 8 ) is divisible by 5 (i.e., ( N equiv 3 pmod{5} )).We start by setting up the system of congruences:[begin{cases}N equiv 5 pmod{8} N equiv 3 pmod{5}end{cases}]Using the Chinese Remainder Theorem, we express ( N ) in terms of one congruence and substitute into the other. Let ( N = 8a + 5 ). Substituting into the second congruence:[8a + 5 equiv 3 pmod{5}]Simplifying, we get:[8a equiv -2 pmod{5} implies 3a equiv 3 pmod{5} implies a equiv 1 pmod{5}]Thus, ( a = 5b + 1 ) for some integer ( b ). Substituting back, we get:[N = 8(5b + 1) + 5 = 40b + 13]The solutions are of the form ( N = 40b + 13 ). Checking values less than 100, we get ( N = 13, 53, 93 ). The largest value is 93.Verification:- For ( N = 93 ): - ( 93 - 5 = 88 ) (divisible by 8) - ( 93 - 8 = 85 ) (divisible by 5)Thus, the maximum number of students is (boxed{93}).]

🔑:10. 93 students10.【Solution】This problem actually involves an integer less than 100, which when reduced by 5 is divisible by 8, i.e., leaves a remainder of 5 when divided by 8, and when reduced by 8 is divisible by 5, i.e., leaves a remainder of 3 when divided by 5, and we need to find its maximum value. 13 leaves a remainder of 5 when divided by 8, and a remainder of 3 when divided by 5. The least common multiple of 8 and 5 is 40, and (13 + 2 times 40 = 93), which is an integer that satisfies the conditions, meaning there are at most 93 students.

🧠:[<think>Alright, let's try to figure out this problem where we need to find all pairs of real numbers (a, b) such that the sequence 10, a, b, ab forms an arithmetic progression. Hmm, arithmetic progression means that the difference between consecutive terms is constant. So, the difference between a and 10 should be the same as between b and a, and the same as between ab and b. Let me write that down step by step.First, in an arithmetic progression (AP), the difference between each term is constant. Let's denote the common difference by d. So, for the sequence 10, a, b, ab to be an AP, the following must hold:1. The difference between the second and first term is d: a - 10 = d2. The difference between the third and second term is d: b - a = d3. The difference between the fourth and third term is d: ab - b = dSince all these differences are equal to d, we can set up equations by equating them. Let's start by equating the first and second differences:a - 10 = b - aThat simplifies to:2a = b + 10=> b = 2a - 10Okay, so we can express b in terms of a. Now let's use the third difference. The difference between the fourth term (ab) and the third term (b) should also be equal to d. So:ab - b = dBut we already know from the second equation that d = b - a. Therefore:ab - b = b - aLet me substitute d here. So replacing d with (b - a) gives:ab - b = b - aLet me rearrange this equation:ab - b - b + a = 0=> ab - 2b + a = 0Factor out terms where possible. Let's see, factor b from the first two terms:b(a - 2) + a = 0Hmm, not sure if that helps immediately. Maybe substitute the value of b from earlier? Since we have b = 2a - 10, let's substitute that into this equation.So replacing b with (2a - 10):(2a - 10)(a - 2) + a = 0Let me expand the first term:(2a - 10)(a - 2) = 2a(a) - 2a(2) -10(a) +10(2)= 2a² - 4a -10a +20= 2a² -14a +20So now the equation becomes:2a² -14a +20 + a = 0=> 2a² -13a +20 = 0Alright, now we have a quadratic equation in terms of a: 2a² -13a +20 = 0. Let's try to solve this quadratic.Quadratic equation is ax² + bx + c = 0. Here, a=2, b=-13, c=20.Discriminant D = b² -4ac = (-13)² - 4*2*20 = 169 - 160 = 9Since the discriminant is positive, we have two real solutions.Solutions are:a = [13 ± √9]/(2*2) = [13 ±3]/4So,a = (13 +3)/4 = 16/4 = 4ora = (13 -3)/4 = 10/4 = 5/2 = 2.5Okay, so we have two possible values for a: 4 and 2.5. Let's find the corresponding b for each using the earlier equation b = 2a -10.First, for a=4:b = 2*4 -10 = 8 -10 = -2Second, for a=2.5:b = 2*(2.5) -10 = 5 -10 = -5So the pairs (a,b) are (4, -2) and (2.5, -5). But wait, let's check if these actually satisfy the original condition of forming an arithmetic progression 10, a, b, ab.First pair: (4, -2). Let's check the sequence:10, 4, -2, (4)*(-2)= -8So sequence is 10, 4, -2, -8Check if it's an AP:Difference between 4 and10 is -6Between -2 and4 is -6Between -8 and -2 is -6Yes, common difference is -6. So that's good.Second pair: (2.5, -5). Let's check the sequence:10, 2.5, -5, (2.5)*(-5)= -12.5Sequence is 10, 2.5, -5, -12.5Check the differences:2.5 -10 = -7.5-5 -2.5 = -7.5-12.5 - (-5) = -12.5 +5 = -7.5Yes, common difference is -7.5. So that also works.Therefore, both pairs are valid. So the solutions are (4, -2) and (5/2, -5). But since the problem says "all pairs of real numbers", we need to present both.Wait, let me just verify once more that when we substituted the values back into ab - b = d, where d = b - a, everything holds. For the first case:a=4, b=-2. Then ab = -8. So ab - b = -8 - (-2) = -8 +2 = -6. And d = b -a = -2 -4 = -6. So that's correct.Second case: a=2.5, b=-5. Then ab = -12.5. ab - b = -12.5 - (-5) = -12.5 +5 = -7.5. And d = b -a = -5 -2.5 = -7.5. Correct again.So both pairs satisfy all the conditions. Therefore, the answer should be these two pairs.Wait, but just to make sure there are no other possibilities. The quadratic gave us two solutions, which we checked. Since the quadratic was derived correctly from the equations, and substitution checks out, I think that's all.Therefore, the final answer is the two pairs: (4, -2) and (5/2, -5).Final AnswerThe pairs are boxed{(4, -2)} and boxed{left( dfrac{5}{2}, -5 right)}.</think>To determine all pairs ((a, b)) of real numbers such that (10, a, b, ab) form an arithmetic progression, we start by noting that the differences between consecutive terms must be constant. Let this common difference be (d). 1. The difference between the second and first term is (d): [ a - 10 = d ]2. The difference between the third and second term is (d): [ b - a = d ]3. The difference between the fourth and third term is (d): [ ab - b = d ]From the first two equations, we equate (a - 10) and (b - a):[a - 10 = b - a implies 2a = b + 10 implies b = 2a - 10]Next, using the third equation and substituting (d = b - a) into (ab - b = d):[ab - b = b - a]Substituting (b = 2a - 10) into the equation:[a(2a - 10) - (2a - 10) = (2a - 10) - a]Simplifying:[(2a - 10)(a - 2) + a = 0]Expanding and simplifying:[2a^2 - 14a + 20 + a = 0 implies 2a^2 - 13a + 20 = 0]Solving the quadratic equation (2a^2 - 13a + 20 = 0) using the quadratic formula:[a = frac{13 pm sqrt{9}}{4} = frac{13 pm 3}{4}]This gives the solutions:[a = 4 quad text{or} quad a = frac{5}{2}]For (a = 4), substituting into (b = 2a - 10):[b = 2(4) - 10 = -2]For (a = frac{5}{2}), substituting into (b = 2a - 10):[b = 2left(frac{5}{2}right) - 10 = -5]Checking both pairs ((4, -2)) and (left(frac{5}{2}, -5right)) confirms that they form an arithmetic progression with the sequence (10, a, b, ab).Thus, the pairs are (boxed{(4, -2)}) and (boxed{left( dfrac{5}{2}, -5 right)}).,<think>Okay, so I need to find all pairs of real numbers (a, b) such that the sequence 10, a, b, ab forms an arithmetic progression. Hmm, let's start by recalling what an arithmetic progression (AP) is. In an AP, the difference between consecutive terms is constant. That means the difference between the second and first term should be the same as the difference between the third and second term, and so on.So, given the sequence 10, a, b, ab, the common difference should be the same each time. Let me denote the common difference as d. Therefore, we can write the following equations:1. a - 10 = d (since the difference between the first and second term is d)2. b - a = d (difference between second and third term)3. ab - b = d (difference between third and fourth term)Since all these differences are equal to d, they should be equal to each other. That means:a - 10 = b - a = ab - bNow, the idea is to set up equations based on these equalities and solve for a and b. Let me first set the first two differences equal:a - 10 = b - aLet me solve this equation. Adding a to both sides and adding 10 to both sides:2a = b + 10So, b = 2a - 10. Got that. So b is expressed in terms of a. Now, I need to use the third equality. Let me set the second difference equal to the third difference:b - a = ab - bLet me substitute b with 2a - 10 here. So:(2a - 10) - a = a*(2a - 10) - (2a - 10)Simplifying the left side first:(2a -10 - a) = a -10Right side:a*(2a -10) - (2a -10) = (2a -10)(a -1)Wait, let me check that again. Let's factor out (2a -10):Right side: (2a -10)(a -1). Hmm, that might be a way to factor it, but let me expand it to be sure.Original right side:a*(2a -10) - (2a -10) = 2a² -10a -2a +10 = 2a² -12a +10So left side is a -10, right side is 2a² -12a +10. Therefore, setting them equal:a -10 = 2a² -12a +10Let's bring all terms to one side:0 = 2a² -12a +10 -a +10Simplify:0 = 2a² -13a +20So quadratic equation: 2a² -13a +20 = 0Now, let's solve this quadratic for a. Let me check if it factors.Looking for two numbers that multiply to 2*20=40 and add up to -13. Hmm. Let's see:-8 and -5: (-8)*(-5)=40 and (-8)+(-5)=-13. Yes, that works.So, we can factor the quadratic as:2a² -8a -5a +20 = 0Grouping terms:(2a² -8a) + (-5a +20) = 02a(a -4) -5(a -4) = 0Factor out (a -4):(2a -5)(a -4) = 0Therefore, solutions are:2a -5 = 0 ⇒ a = 5/2ora -4 = 0 ⇒ a =4So, two possible values for a: 5/2 and 4.Now, let's find the corresponding b using the earlier equation b = 2a -10.First, if a =5/2:b = 2*(5/2) -10 =5 -10 = -5Second, if a=4:b=2*4 -10=8 -10=-2Therefore, the pairs (a,b) are (5/2, -5) and (4, -2).But wait, before concluding, I need to check whether these pairs actually satisfy the original condition that 10, a, b, ab form an arithmetic progression. Sometimes when solving equations, especially with quadratics, we might introduce extraneous solutions. So let me verify both solutions.First pair: (5/2, -5)Sequence: 10, 5/2, -5, (5/2)*(-5) = -25/2Check if this is an arithmetic progression.Compute differences:a -10 = 5/2 -10 = 5/2 -20/2 = -15/2b -a = -5 -5/2 = -15/2ab -b = (-25/2) - (-5) = (-25/2) +10 = (-25/2) +20/2 = (-5/2)Wait, the first two differences are -15/2, then the next difference is (-5/2). So that's not the same. Wait, that's a problem. So this can't be an arithmetic progression. But according to my earlier calculation, it should be. What did I do wrong here?Wait, let's recast the problem. Maybe I made a mistake in assuming the common difference. Let me check each step again.Wait, so if the sequence is 10, a, b, ab, then the differences should be:a -10, b -a, ab -b. So these three differences should be equal.In the first case, when a=5/2 and b=-5:First difference: 5/2 -10 = -15/2Second difference: -5 -5/2 = -15/2Third difference: (-5)*(5/2) - (-5) = (-25/2) +5 = (-25/2 +10/2)= (-15/2). Wait, hold on. Wait, ab is (5/2)*(-5)= -25/2. Then ab -b is (-25/2) - (-5)= -25/2 +5= -25/2 +10/2= -15/2.Oh! Wait, so actually, all three differences are -15/2. So that is an arithmetic progression. But when I first calculated, I thought the third difference was -5/2, but that's a miscalculation. Let me check again:ab is (5/2)*(-5) = -25/2. Then ab - b is -25/2 - (-5) = -25/2 +5 = converting 5 to 10/2, so -25/2 +10/2 = (-15)/2. So yes, the third difference is indeed -15/2. So all differences are equal. Therefore, this is a valid arithmetic progression.Okay, so my first thought was wrong due to miscalculating. So the first pair is valid.Now the second pair: (4, -2)Sequence: 10, 4, -2, 4*(-2)= -8Check the differences:4 -10 = -6-2 -4 = -6-8 - (-2) = -8 +2 = -6So all differences are -6. Therefore, this is a valid arithmetic progression.So both pairs are valid. Therefore, the solutions are (5/2, -5) and (4, -2).Wait, but let me double-check the third term in the first case. The sequence is 10, 5/2, -5, -25/2.Check if the differences are the same:5/2 -10 = -15/2-5 -5/2 = -15/2-25/2 -(-5) = -25/2 +10/2 = -15/2Yes, all differences are -15/2, so that's correct.Therefore, both pairs (5/2, -5) and (4, -2) are valid solutions.But wait, I just want to ensure that there are no other solutions. The quadratic equation had two roots, and both were valid. So that's all. Therefore, these are the only two pairs.But let me just confirm once again by going through the equations.Starting from the original problem: 10, a, b, ab is an arithmetic progression. So:a -10 = b -a = ab - bWe found b =2a -10 from the first two terms. Then substituted into the next equality, leading to quadratic equation, which had two solutions. Both solutions checked out when verified. Therefore, there are exactly two pairs: (5/2, -5) and (4, -2).Hence, these are the solutions.Final AnswerThe pairs are boxed{left( frac{5}{2}, -5 right)} and boxed{(4, -2)}.</think>Given the sequence (10, a, b, ab) which forms an arithmetic progression, we need to determine all pairs ((a, b)) of real numbers. In an arithmetic progression, the difference between consecutive terms is constant. Therefore, we set up the following equations for the common difference (d):1. (a - 10 = d)2. (b - a = d)3. (ab - b = d)From the first two equations, we get:[a - 10 = b - a implies 2a = b + 10 implies b = 2a - 10]Substituting (b = 2a - 10) into the third equation, we get:[b - a = ab - b implies (2a - 10) - a = a(2a - 10) - (2a - 10)]Simplifying both sides:[a - 10 = 2a^2 - 12a + 10]Rearranging terms:[0 = 2a^2 - 13a + 20]Factoring the quadratic equation:[(2a - 5)(a - 4) = 0]Solving for (a), we get:[a = frac{5}{2} quad text{or} quad a = 4]Substituting these values back into (b = 2a - 10):- If (a = frac{5}{2}): [ b = 2 left(frac{5}{2}right) - 10 = 5 - 10 = -5 ]- If (a = 4): [ b = 2(4) - 10 = 8 - 10 = -2 ]Verifying both pairs ((a, b)) in the original sequence:1. For (left( frac{5}{2}, -5 right)): - Sequence: (10, frac{5}{2}, -5, -frac{25}{2}) - Differences: (-frac{15}{2}) each, confirming an arithmetic progression.2. For ((4, -2)): - Sequence: (10, 4, -2, -8) - Differences: (-6) each, confirming an arithmetic progression.Thus, the pairs are (boxed{left( frac{5}{2}, -5 right)}) and (boxed{(4, -2)}).]

🔑:To determine all pairs ((a, b)) of real numbers such that (10, a, b, ab) form an arithmetic progression, we need to use the properties of arithmetic progressions. 1. Identify the common difference: Let the common difference of the arithmetic progression be (d). Then we have: [ a = 10 + d ] [ b = 10 + 2d ] [ ab = 10 + 3d ]2. Express (ab) in terms of (a) and (b): From the above, we know: [ ab = (10 + d)(10 + 2d) ]3. Set up the equation: Since (ab) is also given by (10 + 3d), we equate the two expressions: [ (10 + d)(10 + 2d) = 10 + 3d ]4. Expand and simplify: [ 100 + 20d + 10d + 2d^2 = 10 + 3d ] [ 2d^2 + 30d + 100 = 10 + 3d ] [ 2d^2 + 27d + 90 = 0 ]5. Solve the quadratic equation: Using the quadratic formula (d = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 2), (b = 27), and (c = 90): [ d = frac{-27 pm sqrt{27^2 - 4 cdot 2 cdot 90}}{2 cdot 2} ] [ d = frac{-27 pm sqrt{729 - 720}}{4} ] [ d = frac{-27 pm sqrt{9}}{4} ] [ d = frac{-27 pm 3}{4} ] [ d = frac{-24}{4} quad text{or} quad d = frac{-30}{4} ] [ d = -6 quad text{or} quad d = -7.5 ]6. Find corresponding (a) and (b): - For (d = -6): [ a = 10 + d = 10 - 6 = 4 ] [ b = 10 + 2d = 10 + 2(-6) = 10 - 12 = -2 ] - For (d = -7.5): [ a = 10 + d = 10 - 7.5 = 2.5 ] [ b = 10 + 2d = 10 + 2(-7.5) = 10 - 15 = -5 ]Thus, the pairs ((a, b)) are ((4, -2)) and ((2.5, -5)).The final answer is (boxed{(4, -2) text{ or } (2.5, -5)}).

🧠:[<think>Okay, so I need to find the range of the function y = [x] + 1 when x is between -0.5 and 2.5, not including the endpoints. Let me start by recalling what the floor function does. The floor function, denoted by [x] or sometimes by ⎣x⎦, gives the greatest integer that is less than or equal to x. So, for example, [1.2] would be 1, [3.999] is 3, and [-0.5] would be -1, right?But wait, the problem says the domain is -0.5 < x < 2.5. So x can be any real number greater than -0.5 and less than 2.5. Since the floor function is involved, the key here is to figure out how [x] behaves in different intervals within this domain. Then, once we know [x], we just add 1 to it to get y.Let me break down the domain into intervals where the floor function [x] is constant. The floor function changes its value at integer points. So, within the given domain of -0.5 < x < 2.5, the critical points where [x] changes are at 0, 1, and 2.Therefore, the intervals to consider are:1. -0.5 < x < 02. 0 ≤ x < 13. 1 ≤ x < 24. 2 ≤ x < 2.5Wait, but the original domain is -0.5 < x < 2.5. So, splitting into these intervals based on the integers makes sense. Let me check each interval.Starting with the first interval: -0.5 < x < 0.In this interval, x is negative but greater than -0.5. The floor function [x] for numbers between -0.5 and 0. For example, take x = -0.3. The floor of -0.3 is -1, because -1 is the greatest integer less than or equal to -0.3. Similarly, x = -0.1: [x] = -1. But wait, if x is approaching 0 from the left, like x = -0.0001, then [x] is still -1. So in this entire interval, [x] is -1, so y = [x] + 1 would be -1 + 1 = 0.But let me confirm. The interval is -0.5 < x < 0. The floor function [x] for x between -1 and 0 is -1, right? Wait, but the lower bound here is -0.5, not -1. So for x between -0.5 and 0, since these are numbers greater than -0.5, which is greater than -1, so in that interval, [x] is -1. Because -1 is the greatest integer less than or equal to x, which is still -1 even though x is greater than -0.5. So yes, [x] is -1 here, so y = 0.Next interval: 0 ≤ x < 1.Here, x is from 0 up to but not including 1. For any x in [0,1), the floor function [x] is 0. So y = 0 + 1 = 1. For example, x = 0.5, [x] = 0, so y = 1. Similarly, x = 0.999, [x] = 0, so y = 1. So in this interval, y is 1.Third interval: 1 ≤ x < 2.In this case, x is from 1 up to but not including 2. The floor function [x] is 1 here. So y = 1 + 1 = 2. For example, x = 1.2, [x] = 1, y = 2. x = 1.999, [x] = 1, y = 2. So y is 2 in this interval.Fourth interval: 2 ≤ x < 2.5.Here, x is from 2 up to but not including 2.5. The floor function [x] for x in [2,2.5) is 2. So y = 2 + 1 = 3. For example, x = 2.3, [x] = 2, y = 3. x = 2.499, [x] = 2, y = 3. So in this interval, y is 3.But wait, the original domain is -0.5 < x < 2.5, so we need to check whether all these intervals are included. The first interval is -0.5 < x < 0, then 0 ≤ x < 1, 1 ≤ x < 2, and 2 ≤ x < 2.5. So that covers the entire domain from -0.5 to 2.5.Now, let's list the values of y in each interval:- For -0.5 < x < 0: y = 0- For 0 ≤ x < 1: y = 1- For 1 ≤ x < 2: y = 2- For 2 ≤ x < 2.5: y = 3Therefore, the function y takes on the integer values 0, 1, 2, 3 in these intervals. But let's check if all these values are actually attained.In the first interval (-0.5, 0), since x can approach 0 from the left but doesn't reach 0, y remains 0. Similarly, in the second interval [0,1), y is 1. So 0 is achieved when x is in (-0.5, 0), 1 is achieved when x is in [0,1), 2 when x is in [1,2), and 3 when x is in [2,2.5). So all integers from 0 to 3 are included.But wait, the original question is about the range of y = [x] + 1 for -0.5 < x < 2.5. The range would be the set of all possible y values. So, since in each interval y is 0, 1, 2, 3, but are those all achievable? Let's see.For example, is y = 0 achieved? Yes, when x is in (-0.5, 0), so yes. y = 1 when x is in [0,1), which is part of the domain, so yes. Similarly, y = 2 and y = 3 are achieved. Therefore, the range is the set {0, 1, 2, 3}.But wait, hold on. Let me check the endpoints. The domain is open: -0.5 < x < 2.5, so x cannot be -0.5 or 2.5. But do these affect the range?In the first interval, x approaches -0.5 from the right. Let's say x = -0.5 + ε, where ε is a very small positive number. Then [x] would still be -1, because x is still greater than -0.5 but less than 0. Wait, if x is just above -0.5, say x = -0.499, then [x] is -1. So y = 0. So even approaching -0.5 from the right, y remains 0. Therefore, the lower end of the domain doesn't affect the value here.Similarly, at the upper end, x approaching 2.5 from the left. The interval 2 ≤ x < 2.5, so x approaches 2.5 but doesn't reach it. But in this interval, [x] is 2, so y = 3. So even as x approaches 2.5, y is still 3. Therefore, the open interval doesn't exclude any of these y values.Therefore, the possible values of y are 0, 1, 2, 3. So the range is {0, 1, 2, 3}. But let me check again if there's any overlap or missing values.Wait, but in the first interval, x is between -0.5 and 0, not including -0.5 and 0. So x is greater than -0.5 but less than 0. For all those x, [x] = -1, so y = 0. So y = 0 is included. Then, when x is exactly 0, does that affect anything? Wait, the interval for x is 0 ≤ x < 1, so x = 0 is included here, and [0] = 0, so y = 1. So x = 0 is included, so y = 1 is achieved. Similarly, x = 1 is included in the next interval [1,2), so y = 2. x = 2 is included in [2,2.5), so [2] = 2, y = 3.Therefore, all integers from 0 to 3 are included in the range. So the range is {0, 1, 2, 3}. But I need to represent this as a range. Depending on the problem's requirement, sometimes ranges are given as intervals, but here since the function only takes integer values, the range is the set of these integers.But let me verify once more with specific examples. Take x = -0.4, which is in (-0.5, 0). Then [x] = -1, so y = 0. x = 0, [x] = 0, y = 1. x = 0.5, y = 1. x = 1, y = 2. x = 1.5, y = 2. x = 2, y = 3. x = 2.4, y = 3. So all these y-values are achieved.Therefore, the range is all integers from 0 up to 3. So the answer should be the set {0, 1, 2, 3} or in interval notation if possible. But since these are discrete integers, the range is the set of these four integers.But wait, just to make sure, is there any x in (-0.5, 2.5) that would give a y-value outside of 0,1,2,3? Let's see. The maximum [x] can be in this interval. The maximum x is approaching 2.5, so [x] is 2, so y = 3. The minimum [x] is -1 when x is approaching -0.5 from the right, so y = 0. So there's no way to get below 0 or above 3. So yes, the range is exactly those four integers.Therefore, the answer is the set containing 0,1,2,3. Depending on the notation required, since the problem mentions "the range," and the possible answers in such problems are often given as a list or interval. However, since the function only takes integer values here, the range is the set of integers from 0 to 3 inclusive. So I should write this as {0, 1, 2, 3} or using interval notation if applicable. But in standard math notation for ranges of floor functions, it's typically expressed as the set of integers. So I think {0, 1, 2, 3} is the correct answer.Final AnswerThe range of the function is boxed{{[0, 1, 2, 3]}}.Wait, hold on. The system might expect the range written in another format. Usually, in such cases, the range is presented as a list of integers. But in the final answer, the user instruction says to put the final answer within boxed{}. The example problems I've seen usually have the range as individual numbers in a boxed set. However, sometimes they might present it as interval union. Wait, but since the function y = [x] + 1 here outputs integers, the range is indeed individual integers. But let me check again.Wait, for x in (-0.5, 0), y = 0. So y = 0 is attained. For x in [0,1), y =1. For x in [1,2), y=2. For x in [2,2.5), y=3. So all these integers 0,1,2,3 are attained. Therefore, the range is the set of integers 0,1,2,3.But in the initial problem statement, the answer should be in boxed notation. So perhaps the answer is boxed{{0, 1, 2, 3}}.But wait, the assistant initially thought of putting it as boxed{{[0, 1, 2, 3]}} but that's incorrect because the square brackets inside would denote an interval, but here it's a set of discrete values. So the correct boxed answer should be boxed{{0, 1, 2, 3}}.But let me check once more. Since each interval corresponds to a single integer value, and there are four intervals leading to four integers, the range is those four integers. Therefore, yes, the range is the set {0, 1, 2, 3}, which in boxed form is boxed{{0, 1, 2, 3}}.Final Answerboxed{{0, 1, 2, 3}}</think>To find the range of the function ( y = [x] + 1 ) for ( -0.5 < x < 2.5 ), we need to analyze the behavior of the floor function ([x]) within the given interval. The floor function ([x]) is the greatest integer less than or equal to ( x ).We break down the interval ( -0.5 < x < 2.5 ) into subintervals where the floor function ([x]) is constant:1. Interval ( -0.5 < x < 0 ): - For ( x ) in this interval, ([x] = -1 ). - Therefore, ( y = -1 + 1 = 0 ).2. Interval ( 0 leq x < 1 ): - For ( x ) in this interval, ([x] = 0 ). - Therefore, ( y = 0 + 1 = 1 ).3. Interval ( 1 leq x < 2 ): - For ( x ) in this interval, ([x] = 1 ). - Therefore, ( y = 1 + 1 = 2 ).4. Interval ( 2 leq x < 2.5 ): - For ( x ) in this interval, ([x] = 2 ). - Therefore, ( y = 2 + 1 = 3 ).By examining each subinterval, we determine that the function ( y = [x] + 1 ) takes the values 0, 1, 2, and 3. These values are achieved in their respective intervals, confirming that the range of the function is the set of integers {0, 1, 2, 3}.Thus, the range of the function ( y = [x] + 1 ) for ( -0.5 < x < 2.5 ) is (boxed{{0, 1, 2, 3}}).,<think>Okay, so I need to find the range of the function y equals the floor of x plus one, right? And this is for x values between -0.5 and 2.5. Let me start by recalling what the floor function does. The floor function, [x], gives the greatest integer that is less than or equal to x. For example, [1.2] would be 1, [3.7] would be 3, and [-0.5] would be... hmm, wait, if x is -0.5, then the greatest integer not exceeding -0.5 would be -1, right? Because -1 is less than -0.5, and the next integer, 0, is greater than -0.5. So yeah, [-0.5] is -1.But in the problem, the domain is given as -0.5 < x < 2.5. So x is strictly greater than -0.5 and strictly less than 2.5. So we don't actually include x = -0.5 or x = 2.5. That might be important because endpoints can sometimes change the result, but since they're excluded here, we need to consider open intervals.Alright, so the function is y = [x] + 1. Let's break this down. For any x in the given interval, we first take the floor of x, which gives us an integer, and then add 1 to it. So the value of y will depend on which interval x falls into. Since the floor function is piecewise constant, we can split the domain (-0.5, 2.5) into intervals where [x] is constant, and then add 1 to each of those constants to find the corresponding y values.Let me think about the different intervals within -0.5 to 2.5. The floor function [x] changes its value at integer points. So, the critical points here are x = 0, 1, and 2. Let's split the domain into intervals based on these integers:1. -0.5 < x < 02. 0 ≤ x < 13. 1 ≤ x < 24. 2 ≤ x < 2.5Wait, but x starts at -0.5, which is between -1 and 0. So for the first interval, when x is between -0.5 and 0, [x] is -1, because any number between -0.5 and 0 is greater than -1 but less than 0, so the floor of x is -1. Then, adding 1 gives y = 0.For the next interval, when x is from 0 (inclusive) to 1 (exclusive), [x] is 0, so y = 0 + 1 = 1.Similarly, when x is from 1 (inclusive) to 2 (exclusive), [x] is 1, so y = 1 + 1 = 2.Finally, when x is from 2 (inclusive) to 2.5 (exclusive), [x] is 2, so y = 2 + 1 = 3.Wait, hold on. But let me check these intervals again. For -0.5 < x < 0: since x is greater than -0.5, but less than 0, the floor of x would be -1, because the next integer less than x is -1. For example, take x = -0.3, [x] is -1. Then add 1, so y = 0.For 0 ≤ x < 1: [x] is 0, so y = 1.For 1 ≤ x < 2: [x] is 1, so y = 2.For 2 ≤ x < 2.5: [x] is 2, so y = 3. But wait, x is less than 2.5 here. So x can approach 2.5 but never reaches it. So up to 2.5, but not including it. So at x approaching 2.5 from the left, [x] is still 2. So y remains 3.So then, the y values in each interval would be 0, 1, 2, 3. But let me check if that's accurate. Let me test some values.Take x = -0.4: [x] is -1, so y = 0.x = 0.0: [x] is 0, so y = 1.x = 0.5: [x] is 0, y = 1.x = 1.0: [x] is 1, y = 2.x = 1.9: [x] is 1, y = 2.x = 2.0: [x] is 2, y = 3.x = 2.4: [x] is 2, y = 3.But wait, in the interval definitions above, for 2 ≤ x < 2.5, x is allowed to be 2.0. But since the original domain is -0.5 < x < 2.5, then 2 ≤ x < 2.5 is included. But in the problem statement, x is strictly less than 2.5. So x can be 2.0, 2.1, up to 2.499..., but not 2.5. But even at x=2.499, [x] is 2, so y=3.But let's confirm the endpoints:At x approaching -0.5 from the right (since x > -0.5), say x = -0.49, [x] is -1, so y = 0.At x approaching 0 from the left, x = -0.1, [x] is -1, so y = 0. When x reaches 0, [x] becomes 0, so y = 1.Similarly, at x approaching 1 from the left, x = 0.999, [x] is 0, so y = 1. At x = 1.0, [x] becomes 1, so y = 2.Same with approaching 2 from the left: x = 1.999, [x] is 1, y = 2. At x = 2.0, [x] becomes 2, so y = 3. Then up to x = 2.5, but not including 2.5, so the maximum x is just under 2.5, which still gives [x] = 2, y = 3.Therefore, the possible values of y are 0, 1, 2, 3. But wait, is that the case? Let me check each interval again.First interval: (-0.5, 0) gives y = 0.Second interval: [0, 1) gives y = 1.Third interval: [1, 2) gives y = 2.Fourth interval: [2, 2.5) gives y = 3.Therefore, all these y values are achieved. So the range of the function y = [x] + 1 over -0.5 < x < 2.5 is {0, 1, 2, 3}. But wait, the problem says "the range", which in mathematics usually refers to the set of all output values. However, depending on the context, sometimes it might refer to the interval. But here, since the function is piecewise constant, taking only integer values, the range should be the set of integers from 0 to 3 inclusive.But let me check again. For example, when x is in (-0.5, 0), y = 0. So 0 is included. When x is in [0,1), y = 1, so 1 is included. Similarly, 2 and 3. Therefore, the range is {0,1,2,3}.But wait, the question says "the range of the function y = [x] + 1 for -0.5 < x < 2.5". So the possible y values are 0,1,2,3. Therefore, the range is these four integers. But let me check if all these integers are actually attained.Take x approaching -0.5 from the right: say x = -0.4. Then y = 0. So 0 is in the range.x = 0.0: y = 1. So 1 is in the range.x = 1.0: y = 2. So 2 is in the range.x = 2.0: y = 3. So 3 is in the range. And even x approaching 2.5, like 2.4, y = 3. So 3 is included. So all these values are achieved.But let me check if there are any gaps. For example, is there any x in (-0.5, 2.5) where y could be 0.5? Let's see. Since [x] is always an integer, adding 1 would make y also an integer. So y must be integers. Therefore, the range is only the integers 0,1,2,3. Therefore, the range is the set {0,1,2,3}.But maybe the answer expects it in interval notation or something else? Let me check the problem statement again. It says "the range of the function y = [x] +1 for -0.5 < x < 2.5". The answer should be in boxed notation, so probably as a set of integers. But sometimes range can be expressed in terms of the minimal and maximal values, but since the function is discontinuous, the range isn't an interval. Therefore, the correct answer should be the set containing 0,1,2,3.But let me confirm once more. Suppose x is in (-0.5, 0). Then [x] is -1, so y = 0. So in that interval, y is exactly 0. Then, for x in [0,1), y is 1. Then [1,2), y=2, and [2,2.5), y=3. So all these y values are achieved. Therefore, the range is 0,1,2,3.Therefore, the final answer is {0,1,2,3}, which in boxed notation would be boxed{[0,3]} if they consider it as an interval, but since the function only takes integer values, it's more accurate to write the set. However, in some contexts, ranges for floor functions are sometimes expressed as intervals. Wait, let me think again. For example, if the problem is in a high school setting, sometimes they might expect the range as individual integers. But if in a more advanced context, they might need to write all possible integers. But the problem mentions the Gaussian function, which is another name for the floor function. Then, when you add 1 to the floor function, the outputs are integers. Therefore, the range is those integers. So the answer should be the four integers from 0 to 3 inclusive.But let me check once more for any possible mistakes. For example, when x is approaching 2.5 from the left, the floor of x is 2, so y=3. So 3 is included. When x is approaching -0.5 from the right, say x = -0.4, floor is -1, y=0. So 0 is included. The key here is that each interval between integers shifts the floor value, and adding 1 gives the next integer. Therefore, starting from -0.5, which is just above -1, up to 0, we have floor(x) = -1, then 0 to 1, floor(x)=0, etc.Therefore, the range is indeed the integers 0,1,2,3.But maybe the problem expects it written as an interval from 0 to 3? But since the function doesn't take all values in between, only integers, the range isn't an interval. So the correct answer is the set {0,1,2,3}. But how should this be boxed? In some textbooks, they might write this as [0,3] ∩ ℤ, but I think in the context of the problem, since it's multiple integers, we list them as separate values. But in the format requested, the final answer should be in a box. If the answer is individual integers, perhaps they expect something like boxed{{0,1,2,3}} but I need to check standard conventions. However, looking back at similar problems, often when the range is discrete, they list the set. But in the problem statement, they might prefer interval notation. Wait, but interval notation would imply all real numbers between 0 and 3, which is not the case here. Therefore, the most accurate answer is the set of integers from 0 to 3 inclusive.Alternatively, if the problem is from an Olympiad or competition, sometimes they write the range as the union of intervals, but here since it's individual points, it's better to write the set. But maybe the answer expects it written as individual numbers in boxed notation separated by commas. However, the standard way to present a set in boxed notation is using curly braces. So boxed{{0,1,2,3}}.Alternatively, if the answer is supposed to be in Chinese conventions, but since the user hasn't specified, I think the safest way is to use the set notation.But wait, let me check with examples. For instance, if the floor function is [x], and we have y = [x] +1 over an interval where [x] takes values -1,0,1,2, then y would take 0,1,2,3. So the range is four integers. So the answer is the set containing these four numbers.Alternatively, in some sources, the range might be described as "all integers from 0 to 3 inclusive", which can be written as 0 ≤ y ≤ 3 where y is an integer. But since the problem is in Chinese (the original user might be Chinese), the translation might expect interval notation. Wait, no, in Chinese math education, for discrete ranges, they also use set notation. Therefore, I think the answer is {0,1,2,3} in boxed form.But let me verify once more. If x is in (-0.5,0), [x] = -1, so y = 0. If x is in [0,1), y =1. [1,2): y=2. [2,2.5): y=3. So all these y values are achieved. Since each interval maps to a single y value, the range is the set {0,1,2,3}. Therefore, the correct answer is boxed{{0,1,2,3}}.But sometimes, in multiple-choice or fill-in-the-blank contexts, they might present it as 0,1,2,3 without the set notation. However, given that the problem says "the range", which is a set, it should be presented with set brackets. Therefore, I'll go with boxed{{0,1,2,3}}.Final AnswerThe range of the function is boxed{{0,1,2,3}}.</think>To find the range of the function ( y = [x] + 1 ) for ( -0.5 < x < 2.5 ), we need to consider the behavior of the floor function ([x]) over the given interval. The floor function ([x]) returns the greatest integer less than or equal to ( x ).1. Interval ( -0.5 < x < 0 ): - For ( x ) in this interval, ([x] = -1) because the greatest integer less than ( x ) is -1. - Therefore, ( y = -1 + 1 = 0 ).2. Interval ( 0 leq x < 1 ): - For ( x ) in this interval, ([x] = 0) because the greatest integer less than or equal to ( x ) is 0. - Therefore, ( y = 0 + 1 = 1 ).3. Interval ( 1 leq x < 2 ): - For ( x ) in this interval, ([x] = 1) because the greatest integer less than or equal to ( x ) is 1. - Therefore, ( y = 1 + 1 = 2 ).4. Interval ( 2 leq x < 2.5 ): - For ( x ) in this interval, ([x] = 2) because the greatest integer less than or equal to ( x ) is 2. - Therefore, ( y = 2 + 1 = 3 ).By examining each interval, we determine that the function ( y = [x] + 1 ) takes the values 0, 1, 2, and 3 over the interval ( -0.5 < x < 2.5 ). These values are all integers and there are no other values taken by the function in this interval.Thus, the range of the function ( y = [x] + 1 ) for ( -0.5 < x < 2.5 ) is (boxed{{0, 1, 2, 3}}).]

🔑:1. When -0.5<x<0, the value of the function y=[x]+1 is 0;2. When 0leq x<1, the value of the function y=[x]+1 is 1;3. When 1leq x<2, the value of the function y=[x]+1 is 2;4. When 2leq x<2.5, the value of the function y=[x]+1 is 3;Summarizing the above, the range of the function y=[x]+1 for -0.5<x<2.5 is {0, 1, 2, 3}.Therefore, the answer is boxed{{0, 1, 2, 3}}.

🧠:[<think>Okay, so I have this problem here where I need to figure out which of the options A through E has the largest value. All the options are in the format of 1 minus some decimal number. Let me write them down again to make sure I have them right:A: 1 - 0.1 B: 1 - 0.01 C: 1 - 0.001 D: 1 - 0.0001 E: 1 - 0.00001Hmm, the question is asking which one is the largest. Since all of them start with 1 and subtract a smaller and smaller decimal each time, my initial thought is that the one with the smallest subtracted number would result in the largest value. Let me check that.Starting with option A: 1 minus 0.1 is 0.9. Option B: 1 minus 0.01 is 0.99. Option C: 1 minus 0.001 is 0.999. Option D: 1 minus 0.0001 is 0.9999. Option E: 1 minus 0.00001 is 0.99999.So each subsequent option is subtracting a smaller decimal, which means the result is getting closer and closer to 1. Therefore, the later options (like D and E) should be larger than the earlier ones (A, B, C). Since E subtracts the smallest decimal (0.00001), it should be the closest to 1 and hence the largest value among all options.Wait, let me verify by calculating each one step by step to make sure I didn't make any mistakes in the subtraction.Starting with A: 1 - 0.1 0.1 is the same as 1/10, so subtracting that from 1 gives 9/10, which is 0.9. That's straightforward.B: 1 - 0.01 0.01 is 1/100, so subtracting that from 1 gives 99/100, which is 0.99. That checks out.C: 1 - 0.001 0.001 is 1/1000, so 1 minus that is 999/1000, which is 0.999. Still correct.D: 1 - 0.0001 0.0001 is 1/10,000, so subtracting gives 9999/10,000, which is 0.9999. Yep.E: 1 - 0.00001 0.00001 is 1/100,000, so that leaves 99,999/100,000, which is 0.99999. Definitely.So, when we compare all the results: A: 0.9 B: 0.99 C: 0.999 D: 0.9999 E: 0.99999Each one has more 9s after the decimal point than the previous. Since 0.99999 is greater than 0.9999, which is greater than 0.999, and so on, the order from largest to smallest is E > D > C > B > A.Therefore, the largest value is option E.But wait, just to make sure I didn't overlook anything. All the options are structured similarly, so there's no trick here with negative numbers or anything. Each subtraction is straightforward. The key is recognizing that smaller subtrahends (the numbers being subtracted) lead to larger results. Since 0.00001 is the smallest number being subtracted in option E, that should leave the largest remainder when subtracted from 1.Alternatively, another way to think about it is to compare the subtracted values directly. The problem is asking for the largest result from 1 minus x, where x is each of the decimals given. Since x decreases from A to E, 1 - x must increase from A to E. Therefore, the later the option in the alphabet, the larger the value.So, option E must be the correct answer. I don't see any errors in this reasoning. All steps check out, and each calculation is correct. The logic that subtracting a smaller number from 1 gives a larger result is sound. Therefore, I'm confident that E is the right choice.Final Answerboxed{E}</think>To determine which of the given options has the largest value, we need to evaluate each expression:- A: (1 - 0.1 = 0.9)- B: (1 - 0.01 = 0.99)- C: (1 - 0.001 = 0.999)- D: (1 - 0.0001 = 0.9999)- E: (1 - 0.00001 = 0.99999)Each subsequent option subtracts a smaller decimal from 1, resulting in a value closer to 1. Comparing the results:- A: 0.9- B: 0.99- C: 0.999- D: 0.9999- E: 0.99999The value increases as the subtracted decimal decreases. Therefore, the largest value is from option E, which subtracts the smallest decimal (0.00001) from 1.[boxed{E}],<think>Okay, let's see. The problem is asking which of the options A to E has the largest value. All the options are in the form of 1 minus a small decimal number. Hmm, so each option is subtracting a different decimal from 1. Let me list them out again to make sure I have them right:A: 1 - 0.1B: 1 - 0.01C: 1 - 0.001D: 1 - 0.0001E: 1 - 0.00001So, each subsequent option is subtracting a smaller decimal from 1. That means, as we go from A to E, the number being subtracted gets smaller and smaller. Therefore, the result of 1 minus that number should be getting larger, right? Because subtracting a smaller number from 1 leaves a larger result. Let me check that logic.Take A first: 1 - 0.1. 0.1 is the same as 1/10, so subtracting that from 1 gives 0.9. Then B is 1 - 0.01, which is 0.99. C is 1 - 0.001, which is 0.999. D is 1 - 0.0001, which is 0.9999. And E is 1 - 0.00001, which is 0.99999. So if I compare all these results: 0.9, 0.99, 0.999, 0.9999, 0.99999. Clearly, each one is bigger than the previous. So 0.9 is the smallest, and each time we add another 9 to the decimal, making each subsequent option larger. Therefore, the last option, E, should be the largest.Wait, but let me make sure I didn't misread the options. Sometimes with these multiple-choice questions, there's a trick. Let me verify each calculation again.Starting with A: 1 minus 0.1. Yes, that's 0.9. B: 1 minus 0.01. 0.01 is one-hundredth, so 1 - 0.01 is 0.99. Then C is subtracting a thousandth, so 0.999. D is a ten-thousandth, resulting in 0.9999. E is a hundred-thousandth, so 0.99999. Yep, each time the subtracted number is getting smaller by a factor of 10, so the result is getting closer and closer to 1. Therefore, the largest value among the options is E, since it's the closest to 1. Is there any chance that a larger subtraction could result in a larger number? No, because subtracting a smaller number from 1 will give a result that's closer to 1, hence larger. So, since E has the smallest number subtracted, it should be the largest result. Just to double-check, let's convert each result to decimal form:A: 0.9B: 0.99C: 0.999D: 0.9999E: 0.99999Comparing these, each subsequent number is indeed larger. 0.9 is less than 0.99, which is less than 0.999, and so on. So the order from smallest to largest is A < B < C < D < E. Therefore, E is the largest. I don't see any tricks here. The problem seems straightforward. The key is recognizing that as the subtracted value decreases, the result increases. Since E subtracts the smallest decimal (0.00001), it leaves the largest result. Just to think if there's another way to approach this. Maybe converting all the numbers to fractions? Let's try that.A: 1 - 0.1 = 1 - 1/10 = 9/10 = 0.9B: 1 - 0.01 = 1 - 1/100 = 99/100 = 0.99C: 1 - 0.001 = 1 - 1/1000 = 999/1000 = 0.999D: 1 - 0.0001 = 1 - 1/10,000 = 9999/10,000 = 0.9999E: 1 - 0.00001 = 1 - 1/100,000 = 99,999/100,000 = 0.99999Expressed as fractions, it's clear that each subsequent fraction is closer to 1. The denominators are powers of 10, increasing each time, so the numerators are one less than the denominator. Therefore, each subsequent fraction is larger. So again, E is the largest. Another way: subtract each of the decimals from 1 and see which gives the highest value. Since they are all 1 minus something, the smallest "something" gives the largest result. So since 0.00001 is the smallest number being subtracted, E is the answer.Alternatively, think in terms of place value. The subtracted numbers are in decreasing order of tenths, hundredths, thousandths, ten-thousandths, hundred-thousandths. Each subsequent number is in a smaller decimal place. So subtracting a smaller decimal place from 1 will have a lesser effect on the value, meaning the result is closer to 1. Thus, E is the closest to 1 and hence the largest.Is there any possible way that one of the other options could be larger? For example, maybe a miscalculation in the decimal places? Let me check each subtraction again:A: 1.0 - 0.1. Subtract tenths: 1.0 - 0.1 = 0.9. Correct.B: 1.00 - 0.01 = 0.99. Correct.C: 1.000 - 0.001 = 0.999. Correct.D: 1.0000 - 0.0001 = 0.9999. Correct.E: 1.00000 - 0.00001 = 0.99999. Correct.All these calculations check out. Each result has an additional 9 compared to the previous one, confirming that each is larger. Therefore, there's no error in the individual calculations.Alternatively, perhaps considering negative numbers? But all the options are 1 minus a positive decimal, so all results are positive and less than 1. So the largest among them is the one closest to 1, which is E.Another angle: If we think about the difference from 1. Each option's value is 1 minus a small epsilon. The epsilon for A is 0.1, which is the largest, so A is the farthest from 1. The epsilon for E is 0.00001, the smallest, so E is the closest to 1. Therefore, E is the largest. Yes, that's another way to see it. The closer the number is to 1, the larger it is. So since E is closest to 1, it's the largest.Alternatively, if we subtract each epsilon from 1, the order of the results will be inversely related to the order of the epsilons. Since the epsilons are decreasing (0.1, 0.01, 0.001, etc.), the results must be increasing. Therefore, E is the largest.All these different reasoning paths lead to the same conclusion: option E has the largest value. I can't find any flaws in this reasoning. Therefore, I'm confident that the correct answer is E.Final Answerboxed{E}</think>To determine which of the options A to E has the largest value, we need to evaluate each expression:- A: (1 - 0.1 = 0.9)- B: (1 - 0.01 = 0.99)- C: (1 - 0.001 = 0.999)- D: (1 - 0.0001 = 0.9999)- E: (1 - 0.00001 = 0.99999)Each subsequent option subtracts a smaller decimal value from 1, resulting in a larger number. Comparing the results:- (0.9) (A)- (0.99) (B)- (0.999) (C)- (0.9999) (D)- (0.99999) (E)The values increase as the subtracted decimal decreases. Therefore, the largest value is obtained by subtracting the smallest decimal (0.00001) from 1, which is option E.[boxed{E}]]

🔑:Solution: mathbf{E}In each case a number is subtracted from 1 . The smaller the number we subtract, the larger the answer will be. So the largest value is obtained when the smallest number is subtracted.

🧠:[<think>Okay, so I need to solve this equation: lg(3x² + 12x + 19) - lg(3x + 4) = 1. Hmm, logarithms. Let me recall the logarithm properties. I know that subtracting two logs with the same base is the same as dividing the arguments. So, lg(a) - lg(b) = lg(a/b). Let me check that. Yeah, that's right. So, applying that property here, I can rewrite the left side as a single logarithm.So, the equation becomes lg[(3x² + 12x + 19)/(3x + 4)] = 1. Now, since it's a logarithm base 10, right? Because lg is logarithm base 10. So, if lg(something) = 1, that means that "something" must be 10^1, which is 10. So, (3x² + 12x + 19)/(3x + 4) = 10.Okay, so now I can set up that equation. Let me write that down: (3x² + 12x + 19) divided by (3x + 4) equals 10. To solve for x, I can multiply both sides by (3x + 4) to eliminate the denominator. That gives me 3x² + 12x + 19 = 10*(3x + 4). Let's compute the right side: 10*3x is 30x and 10*4 is 40. So, right side is 30x + 40.So now the equation is 3x² + 12x + 19 = 30x + 40. Let me bring all terms to the left side by subtracting 30x and 40 from both sides. That gives 3x² + 12x + 19 - 30x - 40 = 0. Let me combine like terms. 12x - 30x is -18x, and 19 - 40 is -21. So, the equation simplifies to 3x² - 18x - 21 = 0.Hmm, quadratic equation. Let me see if I can divide both sides by 3 to simplify. Dividing each term by 3: x² - 6x - 7 = 0. Okay, that's simpler. Now, I can try to factor this quadratic. Looking for two numbers that multiply to -7 and add to -6. Let's see, factors of -7: 1 and -7. 1 + (-7) = -6. Yes! So, the quadratic factors as (x - 7)(x + 1) = 0. So, the solutions would be x = 7 and x = -1.But wait, hold on. I need to check if these solutions are valid in the original equation. Because logarithms have domains: the arguments must be positive. So, I need to make sure that both 3x² + 12x + 19 and 3x + 4 are positive when x = 7 and x = -1.Let me check x = 7 first. Compute 3x + 4: 3*7 + 4 = 21 + 4 = 25, which is positive. Then compute 3x² + 12x + 19: 3*(49) + 12*7 + 19 = 147 + 84 + 19 = 250, which is positive. So, x = 7 is okay.Now, x = -1. Compute 3x + 4: 3*(-1) + 4 = -3 + 4 = 1, which is positive. Then compute 3x² + 12x + 19: 3*(1) + 12*(-1) + 19 = 3 -12 + 19 = 10, which is positive. So, x = -1 is also okay.Wait, so both solutions are valid? Hmm. But let me double-check by plugging them back into the original equation.First, x = 7:lg(3*(7)^2 + 12*7 +19) - lg(3*7 +4) = lg(3*49 + 84 +19) - lg(25) = lg(147 +84 +19) - lg25 = lg(250) - lg25. Since 250/25 =10, so lg10 =1, which matches the right side. So, yes, x=7 is correct.Now, x = -1:lg(3*(-1)^2 +12*(-1)+19) - lg(3*(-1)+4) = lg(3*1 -12 +19) - lg(1) = lg(3 -12 +19) - lg1 = lg(10) - 0 = 1 -0 =1. That also works. So, x=-1 is also a valid solution.Wait, so both solutions are valid? But the problem is presented as question 7.096, so maybe expecting two solutions. Hmm. But when I first thought, maybe there's a mistake. Let me check again.Wait, when solving the equation, I had to make sure that when we divided by (3x +4), that (3x +4) is not zero, but x=-4/3 would make it zero. But since x=-1 gives 3x+4=1, which is okay, and x=7 gives 3x+4=25, which is okay, so no division by zero issues.So, both solutions are valid. Therefore, the equation has two real solutions: x=7 and x=-1.But wait, hold on. Wait, the quadratic was x² -6x -7=0, which factors to (x -7)(x +1)=0, so x=7 and x=-1. Correct. So, both solutions are valid.But let me check if maybe when I transformed the original logarithmic equation to the linear equation, I might have introduced extraneous solutions. But in this case, both solutions satisfy the original equation, so they are both valid.Therefore, the answer is x=7 and x=-1.Wait, but the problem is presented as "lg(3x² +12x +19) - lg(3x +4)=1". The original equation. Let me check for x=-1 again. Let's compute 3x +4. 3*(-1) +4 =1, which is positive. 3x² +12x +19=3*(1) + (-12) +19=10, which is positive. So, both logs are defined. Then, lg(10) - lg(1)=1 -0=1. Correct. So, x=-1 is a valid solution.Similarly, for x=7: 3*7 +4=25, positive. 3*49 +84 +19=250, positive. lg250 -lg25=lg(250/25)=lg10=1. Correct.Therefore, both solutions are valid.But wait, maybe in some textbooks or problems, sometimes they only take the principal solution? But in this case, both are mathematically valid. So, I think both x=7 and x=-1 are the solutions.But let me check if there is a mistake in the algebra steps.Original equation: lg(A) - lg(B) =1 => lg(A/B)=1 => A/B=10.So, (3x² +12x +19)/(3x +4)=10.Multiply both sides by (3x +4): 3x² +12x +19 =10*(3x +4).Yes, 3x² +12x +19 =30x +40.Bring all terms to left: 3x² +12x +19 -30x -40=0 => 3x² -18x -21=0. Divide by 3: x² -6x -7=0. Factor: (x-7)(x+1)=0. So x=7 and x=-1. Then check domain, both are valid.Therefore, the solutions are x=7 and x=-1.But wait, maybe the problem expects only one solution? Let me check if when solving, maybe x=-1 makes 3x+4=1, which is positive, but in the original equation, is there a step where maybe dividing by (3x+4) could have caused an issue? No, because when x=-1, (3x+4) is 1, so it's allowed.Therefore, both solutions are correct.But maybe in some contexts, even if mathematically valid, x=-1 is rejected for another reason? Hmm. Let me see. Let me plug x=-1 into the original equation again.lg(3*(-1)^2 +12*(-1)+19) - lg(3*(-1)+4)=lg(3*1 -12 +19) - lg(-3 +4)=lg(10) - lg(1)=1 -0=1. So, yes, it works. So, x=-1 is valid. Therefore, the answer should be both x=7 and x=-1.But let me check if there's a mistake in my factoring. The quadratic is x² -6x -7. To factor, we need two numbers that multiply to -7 and add to -6. So, 7 and -1: 7*(-1)=-7, 7 + (-1)=6. Wait, but we need -6. So, actually, it should be -7 and +1. Wait, no. Let me check again.The quadratic is x² -6x -7. So, coefficient of x is -6, constant term is -7. So, looking for two numbers that multiply to -7 and add to -6. So, the numbers would be -7 and +1 because (-7)(1)=-7 and (-7)+1=-6. Therefore, the factors are (x -7)(x +1). Yes, that's correct. So, x=7 and x=-1. So, correct.Therefore, both solutions are valid, and the answer is x=7 and x=-1.But wait, maybe the original problem is written in base 2? Wait, no, it's specified as lg, which is logarithm base 10. So, all steps are correct.Therefore, the solutions are x=7 and x=-1.Final AnswerThe solutions are boxed{-1} and boxed{7}.</think>Given the equation (lg left(3 x^{2}+12 x+19right)-lg (3 x+4)=1), we start by using the logarithm property that allows us to combine the logarithms into a single expression:[lg left( frac{3x^2 + 12x + 19}{3x + 4} right) = 1]Since (lg(a) = 1) implies (a = 10), we rewrite the equation as:[frac{3x^2 + 12x + 19}{3x + 4} = 10]Next, we eliminate the denominator by multiplying both sides by (3x + 4):[3x^2 + 12x + 19 = 10(3x + 4)]Expanding and simplifying the right side:[3x^2 + 12x + 19 = 30x + 40]We then move all terms to the left side:[3x^2 + 12x + 19 - 30x - 40 = 0]Combining like terms, we get:[3x^2 - 18x - 21 = 0]Dividing the entire equation by 3 simplifies it to:[x^2 - 6x - 7 = 0]Factoring the quadratic equation:[(x - 7)(x + 1) = 0]This gives the solutions (x = 7) and (x = -1). We must check these solutions in the original equation to ensure they do not make the arguments of the logarithms non-positive.For (x = 7):- (3x + 4 = 25) (positive)- (3x^2 + 12x + 19 = 250) (positive)For (x = -1):- (3x + 4 = 1) (positive)- (3x^2 + 12x + 19 = 10) (positive)Both solutions satisfy the original equation. Therefore, the solutions are:[boxed{-1} text{ and } boxed{7}],<think>Okay, so I need to solve the equation: lg(3x² + 12x + 19) - lg(3x + 4) = 1. Hmm, right. Let's start by recalling some logarithm properties. I remember that the difference of two logarithms with the same base is the logarithm of the quotient. So, lg(a) - lg(b) = lg(a/b). That should help simplify the left side of the equation.Applying that property here, the equation becomes:lg[(3x² + 12x + 19)/(3x + 4)] = 1.Now, since it's a logarithm with base 10 (because it's lg, which I think stands for log base 10), I can rewrite this equation in exponential form. Remember that if lg(c) = d, then c = 10^d. So here, that would mean:(3x² + 12x + 19)/(3x + 4) = 10^1 = 10.Okay, so now the equation simplifies to:(3x² + 12x + 19)/(3x + 4) = 10.Now, to solve for x, I can multiply both sides by (3x + 4) to eliminate the denominator. Let's do that:3x² + 12x + 19 = 10*(3x + 4).Let me compute the right side first. 10*(3x + 4) = 30x + 40.So now, the equation is:3x² + 12x + 19 = 30x + 40.Next, bring all terms to one side to set the equation to zero. Subtract 30x and 40 from both sides:3x² + 12x + 19 - 30x - 40 = 0.Simplify the left side by combining like terms. Let's see:12x - 30x = -18x.19 - 40 = -21.So, the equation becomes:3x² - 18x - 21 = 0.Hmm, this is a quadratic equation. Let's see if I can simplify it by dividing all terms by 3 to make the numbers smaller.Dividing each term by 3:x² - 6x - 7 = 0.Okay, that's simpler. Now, solving x² - 6x - 7 = 0. Let's try factoring first. Looking for two numbers that multiply to -7 and add to -6. Hmm, but 7 is a prime number. The factors of -7 would be 1 and -7 or -1 and 7. Let's check:1 + (-7) = -6. Yes! That works. So, the quadratic factors as:(x + 1)(x - 7) = 0.Wait, let me check that. If I expand (x + 1)(x - 7), I get x² -7x + x -7 = x² -6x -7. Exactly. So, the factored form is correct.Setting each factor equal to zero:x + 1 = 0 --> x = -1.x - 7 = 0 --> x = 7.So, the solutions are x = -1 and x = 7.But wait, before finalizing the answer, I need to check if these solutions are valid in the original equation. Because logarithms have domains: the arguments must be positive. So, we need to make sure that both 3x² + 12x + 19 > 0 and 3x + 4 > 0 for each solution.First, check x = -1:Compute 3x + 4 = 3*(-1) + 4 = -3 + 4 = 1 > 0. Good.Compute 3x² + 12x +19 = 3*(1) + 12*(-1) + 19 = 3 -12 +19 = 10 > 0. Good. So, x = -1 is valid.Now, check x = 7:Compute 3x + 4 = 3*7 + 4 = 21 + 4 = 25 > 0. Good.Compute 3x² + 12x +19 = 3*(49) + 12*7 +19 = 147 + 84 +19 = 250 > 0. So, x =7 is also valid.Wait a second, both solutions are valid. So, the answer is x = -1 and x =7. But let me double-check the original equation with these values to be thorough.First, x = -1:Left side: lg(3*(-1)^2 +12*(-1) +19) - lg(3*(-1)+4) = lg(3*1 -12 +19) - lg(-3 +4) = lg(10) - lg(1).lg(10) is 1, lg(1) is 0. So, 1 - 0 =1. Which matches the right side. So, x= -1 is correct.Now, x =7:Left side: lg(3*(7)^2 +12*7 +19) - lg(3*7 +4) = lg(3*49 +84 +19) - lg(21 +4) = lg(147 +84 +19) - lg(25) = lg(250) - lg(25).lg(250) is lg(25*10) = lg(25) + lg(10) = lg(25) +1. So, subtracting lg(25) gives 1. Which matches the right side. So, x=7 is also correct.Therefore, both solutions are valid.But wait, hold on. When I simplified the original equation, I used the logarithm property that allows combining the difference into a quotient. However, we should remember that the domain of the original equation requires that both 3x² +12x +19 >0 and 3x +4 >0. Since both expressions are positive for both solutions, as checked earlier, there's no problem here.But another thing to consider: sometimes when solving logarithmic equations, especially when manipulating them algebraically, we might introduce extraneous solutions. But in this case, both solutions satisfy the original equation, so both are valid.Therefore, the solutions are x = -1 and x =7.But wait, let me check again if the quadratic was solved correctly. Starting from the equation after subtracting 30x +40:3x² +12x +19 -30x -40 =0 => 3x² -18x -21 =0. Dividing by 3: x² -6x -7=0. Factored as (x +1)(x -7)=0. So, x=-1 and x=7. Correct.So, all steps are correct, and both solutions are valid.Therefore, the solutions are x = -1 and x =7.Final AnswerThe solutions are boxed{-1} and boxed{7}.</think>Given the equation (lg left(3 x^{2}+12 x+19right)-lg (3 x+4)=1), we start by using the logarithm property that allows combining the difference into a quotient:[lg left( frac{3x^2 + 12x + 19}{3x + 4} right) = 1]Rewriting this in exponential form, we get:[frac{3x^2 + 12x + 19}{3x + 4} = 10]Multiplying both sides by (3x + 4) to clear the denominator:[3x^2 + 12x + 19 = 10(3x + 4)]Simplifying the right side:[3x^2 + 12x + 19 = 30x + 40]Bringing all terms to one side:[3x^2 + 12x + 19 - 30x - 40 = 0]Combining like terms:[3x^2 - 18x - 21 = 0]Dividing by 3 to simplify:[x^2 - 6x - 7 = 0]Factoring the quadratic equation:[(x + 1)(x - 7) = 0]Solving for (x):[x = -1 quad text{or} quad x = 7]We check the validity of each solution by ensuring the arguments of the logarithms are positive:- For (x = -1): - (3(-1) + 4 = 1 > 0) - (3(-1)^2 + 12(-1) + 19 = 10 > 0)- For (x = 7): - (3(7) + 4 = 25 > 0) - (3(7)^2 + 12(7) + 19 = 250 > 0)Both solutions are valid. Therefore, the solutions are (boxed{-1}) and (boxed{7}).]

🔑:## Solution.Domain of definition: 3 x+4>0, x>-frac{4}{3}.We have lg frac{3 x^{2}+12 x+19}{3 x+4}=1, frac{3 x^{2}+12 x+19}{3 x+4}=10,3 x^{2}-18 x-21=0for 3 x+4 neq 0. From this, x_{1}=-1, x_{2}=7.Answer: -1 ; 7.