🧠:[<think>Okay, let's see. The problem is about a parabola given by the equation y² = 2px, and we need to find the value of p when the directrix is x = -2. The options are A: 2, B:4, C:-2, D:-4. Hmm, I need to recall the standard form of a parabola and how the directrix relates to the parameter p.First, I remember that for a parabola that opens to the right or left, the standard form is y² = 4ax, where the vertex is at the origin, the focus is at (a, 0), and the directrix is the line x = -a. Wait, but in the given equation here, it's y² = 2px. So maybe there's a difference in notation? Let me check.If I compare the two equations: y² = 4ax and y² = 2px. Then, 4a must be equal to 2p, so 4a = 2p, which simplifies to a = p/2. So, the parameter a in the standard form is half of the p given here. Then, the directrix in the standard form is x = -a, so substituting a = p/2, the directrix would be x = -p/2.But in the problem, the directrix is given as x = -2. Therefore, setting -p/2 = -2. Solving for p, we can multiply both sides by -2, so p = (-2)*(-2) = 4? Wait, let me write that step again.If the directrix is x = -2, and according to the standard form, the directrix is x = -a, which equals -p/2. Therefore:-p/2 = -2Multiply both sides by 2:-p = -4Multiply both sides by -1:p = 4So p is 4. Therefore, the answer is B:4.Wait, let me verify again. Let's recall another source. Some references use different notations. For example, in some textbooks, the equation y² = 4ax has a directrix x = -a. So if we have y² = 2px, comparing to y² = 4ax, then 2p = 4a => a = p/2. Then, directrix is x = -a = -p/2. So if the directrix is given as x = -2, then -p/2 = -2 => p = 4. So that seems correct.Alternatively, maybe the student made a mistake here? Let me think. If the equation is y² = 2px, and the directrix is x = -2, then the directrix for a right-opening parabola (if p is positive) is x = -p/2. So setting -p/2 = -2, so p/2 = 2, so p=4. That's correct. So B is the answer. Therefore, I think the correct option is B:4.But wait, let me make sure there's no confusion with the direction. If p is positive, the parabola opens to the right, and the directrix is to the left of the vertex. If p is negative, it opens to the left, and the directrix is to the right. But in this problem, the directrix is x = -2, which is to the left of the vertex (which is at the origin). Therefore, the parabola should open to the right, so p should be positive. So p=4 is positive, which makes sense. So yes, B is correct.Alternatively, if I got confused with another form, like x² = 4py, which is a parabola that opens up or down, but here it's y² = 2px, so definitely horizontal axis. The directrix is vertical line x = -2, which matches the horizontal parabola.Alternatively, maybe there's a formula that directly relates the equation y² = 2px to its directrix. Let me check.In general, for a parabola with a horizontal axis of symmetry, vertex at (h, k), opening to the right or left, the standard form is (y - k)^2 = 4p(x - h). Here, the vertex is at (0, 0), so it's y² = 4p x. Then, the directrix is the line x = h - p, which in this case would be x = 0 - p = -p. Wait, now this is conflicting with the previous notation. Wait, maybe different textbooks use different notations for p.Wait, confusion arises here. Let's clarify. Some references use p as the distance from the vertex to the focus, and then the directrix is on the opposite side at the same distance. So if the focus is at (p, 0), then the directrix is x = -p. But in that case, the standard equation would be y² = 4p x. So if we have y² = 4p x, then directrix is x = -p. So in that case, if the directrix is x = -2, then p = 2. But in our problem, the equation is y² = 2 p x, which is different.So if the standard equation is y² = 4p x, with directrix x = -p, then in our problem, we have y² = 2p x. So comparing to the standard equation 4p x, we can equate 4p = 2p', where p' is the parameter in the given equation. Wait, that might be confusing. Let's define variables properly.Let me use subscripts. Let's say the standard parabola is y² = 4a x, with directrix x = -a. Then, in our problem, we have y² = 2p x. Comparing, 4a = 2p, so a = p/2. Then, the directrix is x = -a = -p/2. Therefore, given the directrix is x = -2, set -p/2 = -2, so p = 4. So that's consistent with the earlier conclusion.Therefore, the correct answer is B:4. So the answer is boxed{B}Final Answerboxed{B}</think>Given the parabola ( y^2 = 2px ) and its directrix equation ( x = -2 ), we need to find the value of ( p ).First, recall the standard form of a parabola that opens to the right or left: ( y^2 = 4ax ). In this form, the directrix is ( x = -a ). Comparing this with the given equation ( y^2 = 2px ), we can equate the coefficients:[ 4a = 2p ]Solving for ( a ), we get:[ a = frac{p}{2} ]The directrix of the given parabola is ( x = -a ). Substituting ( a = frac{p}{2} ), the directrix becomes:[ x = -frac{p}{2} ]Given the directrix is ( x = -2 ), we set up the equation:[ -frac{p}{2} = -2 ]Solving for ( p ):[ frac{p}{2} = 2 ][ p = 4 ]Thus, the value of ( p ) is (boxed{B}).,<think>Okay, so I need to find the value of p for the parabola given by the equation y² = 2px, and they've told me that the directrix is x = -2. The options are A: 2, B: 4, C: -2, D: -4. Let me think about how to approach this.First, I remember that for a parabola in the form y² = 4ax, the directrix is x = -a. Wait, but here the equation is y² = 2px. Maybe I need to compare these two equations to find a relationship between 4a and 2p. Let me write that down.The standard form of a parabola that opens to the right is y² = 4ax. In this case, the directrix is x = -a. Similarly, if the parabola opens to the left, the equation is y² = -4ax, and the directrix is x = a. In the given problem, the equation is y² = 2px. Comparing this to the standard form y² = 4ax, I can equate 2p to 4a. So, 2p = 4a, which simplifies to p = 2a. Hmm, so if I can find the value of a from the directrix, I can find p.The directrix given is x = -2. For the standard parabola y² = 4ax, the directrix is x = -a. So, if the directrix here is x = -2, that would mean -a = -2, so a = 2. Therefore, substituting back into p = 2a, p = 2*2 = 4. So, is the answer B: 4?Wait, let me double-check that. If p = 4, then the equation would be y² = 8x. The standard form would then be y² = 4ax, so 4a = 8, which gives a = 2. Then the directrix should be x = -a = -2, which matches the given directrix. So that seems right.But wait, another thought: maybe I confused the direction. If the equation is y² = 2px, then the parabola opens to the right if p is positive and to the left if p is negative. The directrix is x = -p/2? Wait, no, maybe I got that wrong.Hold on, let me go back to the standard form. The standard form for a parabola that opens right is y² = 4ax. In this case, the directrix is x = -a. So here, in the problem, the equation is y² = 2px. Let's compare to y² = 4ax. Therefore, 2p = 4a. So, a = (2p)/4 = p/2. Therefore, the directrix is x = -a = -p/2.Wait, that's different from what I thought before. So in the problem, the directrix is given as x = -2. Therefore, according to this, x = -p/2 = -2. Solving for p: -p/2 = -2 => p/2 = 2 => p = 4. So that's the same answer as before. So p is 4, which is option B. Therefore, the answer is B.But let me check again to make sure I didn't mix up anything. Let's take p = 4. Then the equation is y² = 8x. The standard form is y² = 4ax, so 4a = 8, so a = 2. Then the directrix is x = -a = -2, which matches the given directrix. So that's correct.Alternatively, if I thought that the directrix is x = -p, then x = -2 would imply p = 2. But according to the standard form, that's not the case. The directrix is related to a, not directly to p. But since in the given equation it's 2p, which is equivalent to 4a, so a is p/2. Therefore, the directrix is x = -a = -p/2.Therefore, solving -p/2 = -2 gives p = 4. So yes, answer B is correct.Wait, maybe another way: for the parabola y² = 2px, the focus is at (p/2, 0) and the directrix is x = -p/2. Is that right? Let me verify.Wait, standard parabola y² = 4ax: focus is at (a, 0), directrix x = -a.So in this problem, equation is y² = 2px. Comparing to y² = 4ax, we get 4a = 2p => a = p/2. Therefore, the focus is at (a, 0) = (p/2, 0) and the directrix is x = -a = -p/2. Therefore, given that the directrix is x = -2, set -p/2 = -2, so p = 4. That confirms again p is 4. So answer is B. Therefore, the answer should be B.Alternatively, if I had a different form, like if the parabola was vertical, but here it's horizontal. Since it's y² = something x, it's a horizontal parabola. So yes, the directrix is vertical line x = something. So everything checks out.I think that's thorough. So the answer is B:4.Final Answerboxed{B}</think>Given the parabola ( y^2 = 2px ) and its directrix equation ( x = -2 ), we need to find the value of ( p ).1. The standard form of a parabola that opens to the right is ( y^2 = 4ax ), where the directrix is ( x = -a ).2. Comparing the given equation ( y^2 = 2px ) with the standard form ( y^2 = 4ax ), we get: [ 2p = 4a implies a = frac{p}{2} ]3. The directrix of the given parabola is ( x = -a ). Substituting ( a = frac{p}{2} ), the directrix becomes: [ x = -frac{p}{2} ]4. Given the directrix ( x = -2 ), we equate: [ -frac{p}{2} = -2 ]5. Solving for ( p ): [ frac{p}{2} = 2 implies p = 4 ]Thus, the value of ( p ) is (boxed{B}).]

🔑:Solution: For the parabola y^2 = 2px with the directrix equation x = -2, the value of p is 4. Therefore, the correct choice is boxed{B}. By using the equation of the directrix of the parabola, we can find the value of p. This question tests the application of the basic properties of parabolas and is a fundamental question.

🧠:[<think>Okay, so let me try to figure out this problem step by step. The problem says: A car was moving at a speed of V. When entering the city, the driver reduced the speed by x%, and upon leaving the city, increased it by 0.5x%. The new speed turned out to be 0.6x% less than the original speed V. We need to find the value of x. Hmm, percentages can be tricky, but let's take it slowly.First, let's parse the problem again to make sure I understand. The car starts at speed V. Then, when entering the city, the speed is reduced by x%. So, the new speed in the city is V minus x% of V. Then, when leaving the city, the speed is increased by 0.5x% of the city speed. After this increase, the final speed is 0.6x% less than the original speed V. So, the final speed is V minus 0.6x% of V. We need to find x such that all these relationships hold.Let me try to write equations for each step. Maybe breaking it down into steps will help.Original speed: VFirst reduction: entering the city, speed is reduced by x%. So, the speed in the city is V * (1 - x/100). Right? Because reducing by x% means you're left with (100 - x)% of the original speed.Then, upon leaving the city, the speed is increased by 0.5x%. Now, here's a critical point: the increase is 0.5x% of the city speed, not the original speed. So, increasing the city speed by 0.5x% would mean multiplying the city speed by (1 + 0.5x/100).So, the speed after leaving the city is: [V * (1 - x/100)] * (1 + 0.5x/100)According to the problem, this final speed is 0.6x% less than the original speed V. So, the final speed is V * (1 - 0.6x/100).Therefore, setting up the equation:[V * (1 - x/100)] * (1 + 0.5x/100) = V * (1 - 0.6x/100)Since V is non-zero, we can divide both sides by V:(1 - x/100) * (1 + 0.5x/100) = (1 - 0.6x/100)Now, we need to solve this equation for x. Let's expand the left side.First, let's denote x/100 as a decimal to make it easier. Let me set k = x/100, so that the equation becomes:(1 - k) * (1 + 0.5k) = 1 - 0.6kBut maybe it's just as easy to work with fractions. Let's expand the left-hand side (LHS):(1 - x/100)(1 + 0.5x/100) = 1*(1) + 1*(0.5x/100) - x/100*(1) - x/100*(0.5x/100)Simplify term by term:First term: 1*1 = 1Second term: 1*(0.5x/100) = 0.5x/100Third term: -x/100*1 = -x/100Fourth term: -x/100*(0.5x/100) = -0.5x²/10000So, combining all terms:1 + 0.5x/100 - x/100 - 0.5x²/10000Simplify the linear terms:0.5x/100 - x/100 = (-0.5x)/100So, LHS becomes:1 - 0.5x/100 - 0.5x²/10000Therefore, the equation is:1 - (0.5x)/100 - (0.5x²)/10000 = 1 - 0.6x/100Subtract 1 from both sides:- (0.5x)/100 - (0.5x²)/10000 = - 0.6x/100Multiply both sides by -1:(0.5x)/100 + (0.5x²)/10000 = 0.6x/100Now, subtract (0.6x)/100 from both sides:(0.5x)/100 - (0.6x)/100 + (0.5x²)/10000 = 0Simplify the terms:(-0.1x)/100 + (0.5x²)/10000 = 0Multiply both sides by 10000 to eliminate denominators:-0.1x * 100 + 0.5x² = 0Calculate -0.1x * 100:-10x + 0.5x² = 0Now, rearrange the terms:0.5x² - 10x = 0Multiply both sides by 2 to eliminate the decimal:x² - 20x = 0Factor out x:x(x - 20) = 0So, x = 0 or x = 20But x = 0 would mean 0% reduction and then 0% increase, which would leave the speed unchanged, but the problem states that the new speed is 0.6x% less than V. If x = 0, then 0.6x% is also 0%, so the speed would still be V. But according to the problem, the driver reduced the speed upon entering the city, so x must be greater than 0. Therefore, the solution is x = 20.Wait, let me check if x = 20 satisfies the original equation.Original equation after dividing by V:(1 - x/100)(1 + 0.5x/100) = 1 - 0.6x/100Plug in x = 20:Left-hand side (LHS):(1 - 20/100)(1 + 0.5*20/100) = (0.8)(1 + 10/100) = 0.8 * 1.1 = 0.88Right-hand side (RHS):1 - 0.6*20/100 = 1 - 12/100 = 0.88Yes, both sides are equal. So, x = 20 is indeed a solution.But wait, x = 0 is also a mathematical solution, but in the context of the problem, x has to be a positive percentage because the driver reduced the speed. So, x = 20 is the valid answer.Let me just go through the problem again to ensure I didn't misinterpret anything.Original speed V.1. Reduced by x% upon entering city: So, new speed is V*(1 - x/100)2. Upon leaving, increased by 0.5x%: So, increase is 0.5x% of the city speed, which is V*(1 - x/100)*(1 + 0.5x/100)3. This new speed is 0.6x% less than V: So, V*(1 - 0.6x/100)So, equation correctly set up. Solved equation, got x = 0 or 20, discard x = 0, answer is x = 20.Yes, that seems right. Let me just check with another value to see if the equation holds. Let's say x = 10. Then LHS would be (1 - 0.1)(1 + 0.05) = 0.9*1.05 = 0.945. RHS would be 1 - 0.06 = 0.94. Not equal, so x = 10 is not a solution. Similarly, x = 20 gives 0.88 on both sides. So, correct.Therefore, the answer is 20.Final AnswerThe value of ( x ) is (boxed{20}).</think>A car was moving at a speed of ( V ). Upon entering the city, the driver reduced the speed by ( x % ), and upon leaving the city, increased it by ( 0.5x % ). The new speed was ( 0.6x % ) less than the original speed ( V ). We need to find the value of ( x ).1. The speed after reducing by ( x % ) is ( V left(1 - frac{x}{100}right) ).2. The speed after increasing by ( 0.5x % ) is ( V left(1 - frac{x}{100}right) left(1 + frac{0.5x}{100}right) ).3. The new speed is ( 0.6x % ) less than ( V ), which is ( V left(1 - frac{0.6x}{100}right) ).Setting up the equation:[V left(1 - frac{x}{100}right) left(1 + frac{0.5x}{100}right) = V left(1 - frac{0.6x}{100}right)]Dividing both sides by ( V ):[left(1 - frac{x}{100}right) left(1 + frac{0.5x}{100}right) = 1 - frac{0.6x}{100}]Expanding the left-hand side:[1 + frac{0.5x}{100} - frac{x}{100} - frac{0.5x^2}{10000} = 1 - frac{0.6x}{100}]Simplifying:[1 - frac{0.5x}{100} - frac{0.5x^2}{10000} = 1 - frac{0.6x}{100}]Subtracting 1 from both sides:[- frac{0.5x}{100} - frac{0.5x^2}{10000} = - frac{0.6x}{100}]Multiplying by -1:[frac{0.5x}{100} + frac{0.5x^2}{10000} = frac{0.6x}{100}]Subtracting (frac{0.6x}{100}) from both sides:[- frac{0.1x}{100} + frac{0.5x^2}{10000} = 0]Multiplying by 10000:[-10x + 0.5x^2 = 0]Factoring:[x(x - 20) = 0]Since ( x ) must be positive, the solution is ( x = 20 ).[boxed{20}],<think>Okay, let's try to tackle this problem step by step. Hmm, so there's a car moving at speed V. Then when entering the city, the driver reduces the speed by x percent. After leaving the city, the driver increases the speed by 0.5x percent. The result is that this new speed is 0.6x percent less than the original speed V. We need to find the value of x.Alright, let's break this down. First, when the car enters the city, the speed is reduced by x%. So the new speed becomes V minus x% of V. Mathematically, that would be V*(1 - x/100). Right? Because a percentage decrease is just multiplying by (1 - percentage/100).Then, upon leaving the city, the driver increases the speed by 0.5x%. So now, the speed is the reduced speed plus 0.5x% of the reduced speed. So that would be the reduced speed multiplied by (1 + 0.5x/100). So putting that together, after leaving the city, the speed is V*(1 - x/100)*(1 + 0.5x/100).Now, according to the problem, this new speed is 0.6x% less than the original speed V. So that means the final speed is V minus 0.6x% of V, which is V*(1 - 0.6x/100). So we have two expressions for the final speed: one from the sequence of speed changes, and one from the comparison to the original speed. Therefore, we can set them equal to each other and solve for x.Let me write this out as an equation:V*(1 - x/100)*(1 + 0.5x/100) = V*(1 - 0.6x/100)Since V is non-zero (the car was moving), we can divide both sides by V to simplify:(1 - x/100)*(1 + 0.5x/100) = (1 - 0.6x/100)Okay, so now we have an equation in terms of x. Let's simplify this equation.First, let's expand the left-hand side (LHS):(1 - x/100)*(1 + 0.5x/100) = 1*(1 + 0.5x/100) - (x/100)*(1 + 0.5x/100)Calculating each term:First term: 1*(1 + 0.5x/100) = 1 + 0.5x/100Second term: (x/100)*(1 + 0.5x/100) = x/100 + 0.5x²/10000So putting it all together:LHS = [1 + 0.5x/100] - [x/100 + 0.5x²/10000]Simplify the terms:1 + (0.5x/100 - x/100) - 0.5x²/100000.5x/100 - x/100 = (-0.5x)/100So LHS becomes:1 - 0.5x/100 - 0.5x²/10000Therefore, the equation is:1 - 0.5x/100 - 0.5x²/10000 = 1 - 0.6x/100Now, subtract 1 from both sides:-0.5x/100 - 0.5x²/10000 = -0.6x/100Multiply both sides by -1:0.5x/100 + 0.5x²/10000 = 0.6x/100Subtract 0.6x/100 from both sides:0.5x/100 + 0.5x²/10000 - 0.6x/100 = 0Combine like terms:(0.5x - 0.6x)/100 + 0.5x²/10000 = 0(-0.1x)/100 + 0.5x²/10000 = 0Multiply all terms by 10000 to eliminate denominators:(-0.1x)*100 + 0.5x² = 0Calculating:-10x + 0.5x² = 0Multiply through by 2 to eliminate the decimal:-20x + x² = 0Rewrite:x² - 20x = 0Factor:x(x - 20) = 0So, the solutions are x = 0 or x = 20.But x = 0 doesn't make sense in the context of the problem because reducing speed by 0% and then increasing by 0% would mean the speed remains V, which wouldn't result in a 0.6x% decrease. Therefore, the valid solution is x = 20.Let me verify this. If x is 20:First, reduce speed by 20%: V*(1 - 20/100) = 0.8VThen, increase speed by 0.5*20 = 10%: 0.8V*(1 + 10/100) = 0.8V*1.1 = 0.88VNow, check if this is 0.6x% less than V. 0.6x = 0.6*20 = 12%. So 12% less than V is V*(1 - 12/100) = 0.88V. Which matches. So yes, x=20 is correct.Therefore, the value of x is 20.Final AnswerThe value of ( x ) is boxed{20}.</think>A car moving at speed ( V ) reduces its speed by ( x % ) upon entering the city, resulting in a speed of ( V left(1 - frac{x}{100}right) ). Upon leaving the city, the speed is increased by ( 0.5x % ), resulting in a speed of ( V left(1 - frac{x}{100}right) left(1 + frac{0.5x}{100}right) ). This new speed is ( 0.6x % ) less than the original speed ( V ), which is ( V left(1 - frac{0.6x}{100}right) ).Setting these two expressions equal gives:[V left(1 - frac{x}{100}right) left(1 + frac{0.5x}{100}right) = V left(1 - frac{0.6x}{100}right)]Dividing both sides by ( V ):[left(1 - frac{x}{100}right) left(1 + frac{0.5x}{100}right) = 1 - frac{0.6x}{100}]Expanding the left-hand side:[1 + frac{0.5x}{100} - frac{x}{100} - frac{0.5x^2}{10000} = 1 - frac{0.6x}{100}]Simplifying:[1 - frac{0.5x}{100} - frac{0.5x^2}{10000} = 1 - frac{0.6x}{100}]Subtracting 1 from both sides:[- frac{0.5x}{100} - frac{0.5x^2}{10000} = - frac{0.6x}{100}]Multiplying by -1:[frac{0.5x}{100} + frac{0.5x^2}{10000} = frac{0.6x}{100}]Subtracting (frac{0.6x}{100}) from both sides:[frac{0.5x - 0.6x}{100} + frac{0.5x^2}{10000} = 0]Simplifying:[- frac{0.1x}{100} + frac{0.5x^2}{10000} = 0]Multiplying by 10000:[-10x + 0.5x^2 = 0]Multiplying by 2:[-20x + x^2 = 0]Factoring:[x(x - 20) = 0]Thus, ( x = 0 ) or ( x = 20 ). Since ( x = 0 ) is not practical, the solution is ( x = 20 ).The value of ( x ) is boxed{20}.]

🔑:Answer: 20. Solution. The condition of the problem means that the equation is satisfiedvleft(1-frac{x}{100}right)left(1+frac{0.5 x}{100}right)=vleft(1-frac{0.6 x}{100}right) Leftrightarrowleft(1-frac{x}{100}right)left(1+frac{x}{200}right)=1-frac{3 x}{500} Leftrightarrow frac{x^{2}}{20000}=frac{3 x}{500}-frac{x}{200} .Leftrightarrow x=0 ; x=20 text {. The value } x=0 text { contradicts the condition. Therefore, } x=20 text {. }

🧠:[<think>Okay, so I need to figure out for which values of x the product (x - 4)(x² - 9x + 20) is positive. Hmm, let's start by understanding what the problem is asking. They want the intervals where the product of these two expressions is greater than zero. That means I have to find where both factors are positive or both are negative. First, maybe I should factor the quadratic expression to make things simpler. Let me look at x² - 9x + 20. To factor this, I need two numbers that multiply to 20 and add up to -9. Let's see, factors of 20: 1 and 20, 2 and 10, 4 and 5. The pair 4 and 5 adds up to 9. Since the middle term is -9x, both numbers should be negative. So, x² - 9x + 20 factors into (x - 4)(x - 5). Let me check that: (x - 4)(x - 5) = x² -5x -4x +20 = x² -9x +20. Yes, that works.So the original expression becomes (x - 4)(x - 4)(x - 5). Wait, that's (x - 4)²(x - 5). So the product is (x - 4) squared multiplied by (x - 5). Now, since (x - 4) squared is always non-negative (because any real number squared is non-negative), the sign of the entire product depends on the sign of (x - 5) and whether (x - 4)² is zero or positive. But we're interested in when the product is positive. So, let's note that (x - 4)² is zero when x = 4, and positive otherwise. So, the product (x - 4)²(x - 5) will be zero when x = 4 or x = 5 (since x = 5 makes (x - 5) zero). But we want the product to be positive. So, since (x - 4)² is always non-negative, the sign of the entire expression depends on (x - 5). However, (x - 4)² is positive except when x = 4, where it's zero. Therefore, the product will be positive when (x - 5) is positive, except at x = 4 where the product is zero. So, if x - 5 > 0, which is x > 5, then the entire product is positive. But wait, is that the only interval?Wait, hold on. Let me think again. If (x - 4)² is positive except at x = 4, and (x - 5) is positive when x > 5 and negative when x < 5. So, when x > 5, both factors (x - 4)² and (x - 5) are positive (since (x - 4)² is always non-negative but positive except at x=4), so their product is positive. When x < 5, (x - 5) is negative, but (x - 4)² is still positive (except at x=4, where it's zero). Therefore, the product would be negative when x < 5 (since positive times negative is negative), except when x = 4, where the product is zero. So, does that mean that the product is positive only when x > 5? But wait, what about between x=4 and x=5? Let's check a test value. Let's take x=4.5. Then (x - 4)² is (0.5)² = 0.25, which is positive. (x - 5) is -0.5, so the product is 0.25*(-0.5) = -0.125, which is negative. So between 4 and 5, the product is negative. What about at x=3? (3 -4)^2=1, (3 -5)= -2, so product is 1*(-2)= -2. Still negative. At x=6, (6-4)^2=4, (6 -5)=1, product=4*1=4, positive. So indeed, only when x>5, the product is positive. But wait, but we also need to check whether (x -4)^2 can affect the sign. Since it's squared, it's always non-negative, so the only time the product is zero is at x=4 or x=5. So, if I set (x -4)^2(x -5) > 0, then since (x -4)^2 is non-negative, the inequality holds when (x -5) is positive and (x -4)^2 is not zero. Therefore, x -5 >0 and x ≠4. But since x>5 already implies x ≠4, the solution is x>5. But let me double-check. Let's consider the original problem: (x -4)(x² -9x +20). We factored that as (x -4)(x -4)(x -5). So it's (x -4)^2(x -5). Therefore, the critical points are x=4 and x=5. So the intervals to test are (-infty,4), (4,5), and (5,infty). In (-infty,4): Let's take x=0. (0 -4)^2 =16, (0 -5)= -5. 16*(-5)= -80 <0. So negative here. In (4,5): Take x=4.5. As before, product is negative. In (5, infty): Take x=6. Product is positive. At x=4: Product is 0. At x=5: Product is 0. So the only interval where the product is positive is x>5. Therefore, the solution is x>5. But wait, hold on. Let me check if there's any other interval where the product could be positive. Since (x -4)^2 is always positive except at x=4, the sign of the product is entirely determined by (x -5). Therefore, when (x -5) is positive (x>5), the product is positive, and when (x -5) is negative (x<5), the product is negative. But when x=5, the product is zero, and at x=4, the product is zero. Therefore, the answer should be x>5. So, written in interval notation, that's (5, ∞). Wait, but let me make sure I didn't skip any steps. Let me consider the original problem again. If I didn't factor the quadratic, would I get the same result?Original expression: (x -4)(x² -9x +20). If I didn't factor the quadratic, I would still need to find when the product is positive. To do that, I should find the roots of each factor and test the intervals. First factor: x -4. Root at x=4. Second factor: x² -9x +20. Let's find its roots. Using quadratic formula: x = [9 ± sqrt(81 -80)] / 2 = [9 ±1]/2. So x=(9+1)/2=10/2=5, and x=(9-1)/2=8/2=4. So the roots are x=4 and x=5. So the quadratic factors as (x -4)(x -5), which is the same as before. Therefore, the original expression is (x -4)(x -4)(x -5). So the roots are x=4 (double root) and x=5. So to determine the intervals where the product is positive, we can consider the number line divided by the roots x=4 and x=5. But since x=4 is a double root, the sign of the expression may not change there. Let's see. Let me set up a sign chart. Critical points at x=4 and x=5. For x <4: All three factors (x -4) [twice] and (x -5) are negative. But since (x -4) is squared, two negatives multiplied give a positive, then times (x -5) which is negative. So positive times negative is negative. Between x=4 and x=5: (x -4) is positive (squared, so positive), and (x -5) is negative. So positive times negative is negative. For x >5: (x -4) is positive (squared) and (x -5) is positive. So positive times positive is positive. Therefore, indeed, the product is positive only when x >5. Therefore, the solution is all real numbers x where x >5. But let me just verify with another test point. Let's take x=6: (6-4)=2, (6² -9*6 +20)=36 -54 +20=2. So (2)(2)=4, which is positive. Take x=4.5: (4.5-4)=0.5, (4.5² -9*4.5 +20)=20.25 -40.5 +20= -0.25. So (0.5)(-0.25)= -0.125, which is negative. Take x=3: (3-4)= -1, (3² -9*3 +20)=9 -27 +20=2. So (-1)(2)= -2, which is negative. So yes, only when x>5 is the product positive. Therefore, the answer is x >5. In interval notation, that's (5, ∞). But let me check if there's a mistake in the factoring. The quadratic was x² -9x +20, which factors into (x -4)(x -5). So the original product is (x -4)*(x -4)*(x -5) = (x -4)^2*(x -5). Since (x -4)^2 is non-negative, the sign is determined by (x -5). So when x >5, the whole expression is positive. When x <5, it's negative. At x=4, it's zero. At x=5, it's zero. Therefore, the solution is x >5. But just to make sure, let's think if there's any case where (x -4)^2 could be zero. That's only when x=4, but there the product is zero, which is not positive. So x=4 is excluded. Therefore, all values of x greater than 5 make the product positive. I think that's solid. So the answer is x >5, which in boxed form is boxed{(5, infty)}.Final AnswerThe values of ( x ) for which the product is positive are boxed{(5, infty)}.</think>To determine the values of ( x ) for which the product ( (x-4)(x^2 - 9x + 20) ) is positive, we start by factoring the quadratic expression. The quadratic ( x^2 - 9x + 20 ) factors into ( (x - 4)(x - 5) ). Thus, the original expression becomes:[(x - 4)(x - 4)(x - 5) = (x - 4)^2(x - 5)]Next, we analyze the sign of the product ( (x - 4)^2(x - 5) ). Since ( (x - 4)^2 ) is always non-negative (it is zero when ( x = 4 ) and positive otherwise), the sign of the product depends on the factor ( (x - 5) ).- When ( x > 5 ), ( (x - 5) ) is positive, and since ( (x - 4)^2 ) is positive, the product is positive.- When ( x < 5 ), ( (x - 5) ) is negative, and even though ( (x - 4)^2 ) is positive, the product is negative.- At ( x = 4 ) and ( x = 5 ), the product is zero.Testing intervals around the critical points ( x = 4 ) and ( x = 5 ) confirms that the product is positive only when ( x > 5 ).Thus, the values of ( x ) for which the product is positive are:[boxed{(5, infty)}],<think>Okay, so I need to figure out for which values of x the product (x - 4)(x² - 9x + 20) is positive. Hmm, let's start by understanding what the problem is asking. They want the set of all real numbers x where this product is greater than zero. That means I have to find the intervals where the expression is positive. First, maybe I should factor the quadratic expression in the second term to simplify the product. The quadratic is x² - 9x + 20. Let me try to factor that. To factor a quadratic of the form x² + bx + c, I need two numbers that multiply to c and add up to b. In this case, c is 20 and b is -9. So I need two numbers that multiply to 20 and add up to -9. Let's see: the factors of 20 are 1 and 20, 2 and 10, 4 and 5. Hmm, 4 and 5 add up to 9. If both are negative, they would add up to -9. So, -4 and -5. Yeah, that works. So, x² - 9x + 20 factors to (x - 4)(x - 5). Let me check: (x - 4)(x - 5) = x² -5x -4x +20 = x² -9x +20. Yep, that's correct.So the original product becomes (x - 4)(x - 4)(x - 5). Wait, that's (x - 4) squared times (x - 5). So, the expression is (x - 4)^2 (x - 5). Now, since (x - 4)^2 is a square, it's always non-negative, right? Because any real number squared is non-negative. So the sign of the entire expression depends on the sign of (x - 5), except when (x - 4)^2 is zero, which would make the entire expression zero. But since we want the product to be positive, we need to consider when (x - 4)^2 is positive and (x - 5) is positive, or when (x - 4)^2 is positive and (x - 5) is negative. Wait, but (x - 4)^2 is non-negative. So, if (x - 4)^2 is zero, the product is zero, which we don't want. So the product is positive only when (x - 4)^2 is positive and (x - 5) is positive. Because (x - 4)^2 can't be negative. So the entire product's sign is determined by (x - 5). Wait, is that right? Let me think again. If (x - 4)^2 is positive, which is true for all x except x=4, then the sign of the entire expression is the same as the sign of (x - 5). So when (x - 5) is positive, the product is positive; when (x - 5) is negative, the product is negative. And at x=5, the product is zero. At x=4, the product is zero as well. So, to find where the product is positive, we need to find where (x - 5) is positive, but also considering that x ≠ 4 because at x=4 the product is zero, which is not positive. So (x - 5) > 0 implies x > 5. But we need to exclude x=4, but since x=4 is less than 5, in the interval x > 5, x=4 is not included. So the solution is x > 5. But wait, let me double-check by testing intervals. The critical points are x=4 and x=5. Let's divide the number line into intervals based on these points: (-∞, 4), (4, 5), and (5, ∞). Then test each interval. First interval: x < 4. Let's pick x=3. Then (3 - 4)^2 = ( -1)^2 = 1, which is positive. (3 - 5) = -2, which is negative. So positive times negative is negative. So the product is negative here. Second interval: 4 < x <5. Let's pick x=4.5. Then (4.5 -4)^2 = (0.5)^2 =0.25, positive. (4.5 -5) = -0.5, negative. So product is positive * negative = negative. Third interval: x >5. Let's pick x=6. (6 -4)^2 = 4, positive. (6 -5)=1, positive. Product is positive*positive=positive. At x=4: product is zero. At x=5: product is zero. So the product is positive only when x>5. Therefore, the solution is x>5. Wait, but let me make sure I didn't miss anything. The original expression is (x -4)(x² -9x +20). If I didn't factor correctly or made a mistake in combining the terms... But no, x² -9x +20 factors into (x-4)(x-5), so the product is (x-4)^2(x-5). So that's correct. Alternatively, another way to approach this is to consider the original expression as (x -4)(x² -9x +20) and find where it is positive. To do that, we can find the roots of each factor and test the intervals. First, find the roots of (x -4): root at x=4. Then find roots of (x² -9x +20). As before, x=4 and x=5. So overall, the roots are x=4 (double root) and x=5. Wait, but actually, the quadratic factors into (x -4)(x -5), so the entire expression is (x -4)*(x -4)*(x -5). Therefore, the roots are x=4 (with multiplicity 2) and x=5 (with multiplicity 1). So when constructing the sign chart, the critical points are at x=4 and x=5. The multiplicity affects whether the sign changes or not. Since x=4 is a double root, the sign does not change when passing through x=4. The sign changes at x=5 because it's a single root. So, let's set up intervals:1. x <4: Let's test x=3. Then (3 -4)= -1, (3 -4)= -1, (3 -5)= -2. So product is (-1)*(-1)*(-2)= -2, which is negative. 2. Between 4 and5: Let's test x=4.5. (4.5 -4)=0.5, (4.5 -4)=0.5, (4.5 -5)= -0.5. So product is (0.5)(0.5)(-0.5)= -0.125, which is negative. 3. x >5: Let's test x=6. (6 -4)=2, (6 -4)=2, (6 -5)=1. Product is 2*2*1=4, positive. Therefore, the product is positive only when x>5. So the answer is x>5, which in interval notation is (5, ∞). But wait, just to make sure, maybe I should check another value in the interval x>5. Let's pick x=10. Then (10 -4)^2*(10 -5)=6^2*5=36*5=180, which is positive. And another in x<4, say x=0: (0 -4)^2*(0 -5)=16*(-5)=-80, which is negative. Between 4 and5, like x=4.5, as before, negative. Therefore, yes, the product is positive only when x>5. So the solution is all real numbers greater than 5. I think that's it. So the answer is x >5, or in box notation, boxed{(5, infty)}.Final AnswerThe product is positive for all real numbers greater than 5, so the solution is boxed{(5, infty)}.</think>To determine for which values of ( x ) the product ((x - 4)(x^2 - 9x + 20)) is positive, we start by factoring the quadratic expression. The quadratic ( x^2 - 9x + 20 ) factors into ( (x - 4)(x - 5) ). Thus, the original expression becomes:[(x - 4)(x - 4)(x - 5) = (x - 4)^2(x - 5)]Next, we analyze the sign of this product. Since ((x - 4)^2) is a square term, it is always non-negative and is zero only when ( x = 4 ). The term ((x - 5)) determines the sign of the product because ((x - 4)^2) is non-negative. We consider the critical points ( x = 4 ) and ( x = 5 ) to divide the number line into intervals and test the sign of the product in each interval:1. Interval ( (-infty, 4) ): - Choose ( x = 3 ): ((3 - 4)^2(3 - 5) = (-1)^2(-2) = -2 ) (negative)2. Interval ( (4, 5) ): - Choose ( x = 4.5 ): ((4.5 - 4)^2(4.5 - 5) = (0.5)^2(-0.5) = -0.125 ) (negative)3. Interval ( (5, infty) ): - Choose ( x = 6 ): ((6 - 4)^2(6 - 5) = (2)^2(1) = 4 ) (positive)At ( x = 4 ) and ( x = 5 ), the product is zero, which is not positive. Therefore, the product is positive only in the interval ( (5, infty) ).[boxed{(5, infty)}]]

🔑:The product can also be written as:(x-4)(x-4)(x-5)or(x-4)^{2}(x-5)This expression is positive when x>5.Number of solutions: 24.

🧠:[<think>Okay, let's tackle this problem step by step. We have four propositions and we need to determine how many of them are correct. The options are from 0 to 4, so I need to check each one carefully. Let me go through each proposition one by one.Starting with Proposition (1): If a corner of an n-sided paper piece is cut off with scissors (n ≥ 4 and the cutting line is a straight line), the remaining paper piece is an (n-1)-sided or (n+1)-sided polygon.Hmm, so cutting off a corner of a polygon. Let me visualize this. If you have, say, a square (n=4) and you cut off one corner with a straight line. What happens? The original corner is a vertex, and cutting it off with a straight line would replace that corner (vertex) with a new edge. So instead of the original two edges meeting at a vertex, we now have a new edge connecting the two adjacent edges. So for a square, cutting off a corner would replace one vertex with an edge, so the number of sides would increase by 1? Wait, no. Wait, the square has four sides. If you cut off a corner, you remove one vertex but add two edges? Wait, no. Let me think again.Wait, when you cut off a corner, you're removing a vertex but replacing it with a new edge. So each original vertex is a corner where two edges meet. If you cut off the vertex with a straight line, you are effectively replacing that vertex with a new edge. So each cut removes one vertex but adds two new edges? Wait, that doesn't seem right. Wait, suppose you have a polygon, and you cut off a corner. The original polygon has n sides. When you cut a corner, you're cutting from one edge to another adjacent edge. So, for example, in a square, cutting from one edge to the next edge over the corner. That would replace the original two edges (each adjacent to the corner) with a single new edge. Wait, but in reality, cutting a corner with a straight line would remove one vertex and replace two edges with one edge. So the number of sides would decrease by 1. Wait, but if you cut off a corner, like cutting diagonally across the corner, you actually split the original two edges into two parts each, but connected by the new edge. Wait, maybe I need to draw a mental picture.Wait, original polygon: let's say a square. Each corner is a 90-degree angle. If you cut off a corner with a straight line, you are removing that corner and replacing it with a new edge. So instead of two edges meeting at a vertex, you now have a new edge connecting the two adjacent edges. So in the square, each original edge is adjacent to two vertices. Cutting off one vertex would remove that vertex and replace the two edges connected to it with a single edge. Therefore, the number of sides would decrease by one. Wait, but a square has four sides. If you cut off one corner, you have a pentagon? Wait, that can't be. Wait, if you take a square and cut off a corner, the resulting shape is a pentagon. Wait, let me count. Original square: four sides. Cut off a corner: the cut goes from one side to another, adjacent side. So each side adjacent to the corner is now divided into two parts by the cut. Therefore, each original side is split, and the cut creates a new side. So the total number of sides becomes 4 - 1 + 2 = 5? Wait, maybe. Wait, original square has four sides. Each corner is where two sides meet. If you cut off the corner, you remove a part of each of those two sides and replace them with a single new side. So instead of two sides (each partially remaining) plus the new side, so total sides: original four sides minus two sides (each being cut) plus three sides (the remaining parts of the original two sides plus the new cut side). Wait, this is getting confusing. Maybe a better approach.Alternatively, Euler's formula? Wait, maybe not necessary. Let's consider specific examples. If you have a square (4 sides) and you cut off one corner. The result is a pentagon (5 sides). Wait, that's actually correct. When you cut a corner off a square, you add a new edge where the corner was, so each cut adds a side. Wait, but if you cut a corner, you are replacing a vertex (a corner) with an edge. So each vertex you cut off, you replace it with an edge, so the number of sides increases by 1. So n becomes n + 1. So for a square, cutting a corner makes it a pentagon. But wait, another thought. Suppose you have a polygon, and you cut off a corner in such a way that the cut goes from one side to another non-adjacent side. Wait, but the problem states that the cutting line is a straight line. But in a polygon, the sides are adjacent. So cutting a corner would have to go from one side to an adjacent side. Wait, maybe if you cut a corner, the line goes from one side to the next, so effectively, you are truncating the corner.Alternatively, if you have a polygon and you cut off a vertex (corner), you are essentially replacing that vertex with a new edge. So each vertex removed is replaced by an edge. So if you start with n sides, cutting off a vertex (which is the meeting point of two sides) would replace that vertex with a new edge. Therefore, the number of sides would increase by 1. So n becomes n + 1.But wait, let's test with triangle. If you have a triangle (3 sides) and cut off a corner, you get a quadrilateral (4 sides). Similarly, square becomes pentagon. So seems like cutting a corner increases the number of sides by 1. But the proposition says that it's either (n - 1) or (n + 1). So according to my examples, it becomes n + 1. So why does the proposition say it could be n - 1?Wait, maybe the direction of the cut. If you cut off a corner such that the cut is between two non-adjacent sides? But in a polygon, sides are adjacent. Wait, maybe if you have a star-shaped polygon or something else? But the problem states it's an n-sided paper piece, which is a polygon with n sides, and n ≥ 4. So maybe if the cut is along a line that connects two existing vertices, effectively removing a side?Wait, no. If you cut off a corner, it's a straight line. Suppose you have a polygon, and you cut from one vertex to another, but the problem says "a corner of an n-sided paper piece is cut off with scissors... the cutting line is a straight line". So cutting off a corner would mean removing the vertex by making a straight cut. If the cut is between two adjacent sides, replacing the vertex with a new side. So that would increase the number of sides by 1, as in the examples. But maybe in some cases, it could decrease?Wait, maybe if the cut is such that it removes a side? For example, if you have a polygon and you make a cut that goes from one side to another, but not adjacent. Wait, but in a convex polygon, any straight cut can only intersect two sides. If the polygon is convex, cutting off a corner (vertex) would always increase the number of sides by 1. If the polygon is concave, maybe the cut could connect two non-adjacent sides? But even then, cutting off a vertex would still replace it with a new edge. Wait, perhaps I need to think again.Wait, let's take a different example. Suppose we have a rectangle (4 sides). If we cut off a corner, which is a vertex, by making a diagonal cut from one adjacent side to the other. So cutting from the top side near the top-right corner to the right side near the top-right corner. Wait, but in that case, you are replacing the 90-degree corner with a diagonal edge, so the number of sides becomes 5. So original 4 sides, cutting off a corner, adding one side, total 5.Alternatively, suppose you have a polygon and you cut off a corner in such a way that the cut line goes from a side to the same side again. Wait, that's not possible with a straight line. A straight cut has to go from one side to another side.Wait, maybe if the cut is along an existing edge? But then you're not cutting off a corner. Hmm. Alternatively, maybe if the cut is such that it removes two sides? Wait, in a polygon, each side is connected to two vertices. If you make a straight cut from one side to another, but in such a way that the cut removes two sides? For example, in a square, if you cut from the midpoint of one side to the midpoint of the opposite side, but that would divide the square into two pieces, not cutting off a corner.Wait, the problem states "a corner of an n-sided paper piece is cut off with scissors". So it's a single cut that removes the corner, resulting in a single remaining piece. So in that case, the cut must start and end on two adjacent sides, right? Because the corner is where two sides meet. So to cut off the corner, you make a straight line from one side to the adjacent side, removing the corner. Therefore, that cut replaces the original vertex (corner) with a new edge. Therefore, the number of sides would be n + 1. Because you had n sides, each original corner connected two sides. Cutting off the corner would replace the two sides adjacent to the corner with two truncated sides and a new edge between them. Wait, so original n sides, each side adjacent to two corners. If you cut off one corner, you replace two sides (each adjacent to the corner) with two sides minus the truncated parts, plus the new edge. Wait, but actually, when you cut a corner, you're cutting from one side to the next, so each original side is split into two parts by the cut. But since we're removing the corner, we discard the parts beyond the cut. So the remaining polygon would have the original sides minus the parts beyond the cut, plus the new edge. Therefore, for each corner cut off, two sides are truncated (each losing a segment) and one new side is added. Therefore, the total number of sides becomes (n - 2) + 1 = n - 1. Wait, this contradicts my earlier thought.Wait, let's take an example. Take a square (4 sides). Cut off one corner. How many sides does the resulting shape have? If I make a straight cut from one side to the adjacent side at a corner, removing the corner. The original square has four sides. After cutting off the corner, each of the two sides adjacent to the corner are now shorter, and there's a new side where the cut was made. So instead of four sides, we now have the two original sides (each truncated) plus the new cut side, and the other two sides remain the same. Wait, no. Wait, in the square, cutting off one corner affects two sides. Each of those sides is now split into two parts: the part before the cut and the part after. But since we're removing the corner, we discard the parts beyond the cut. So each of the two sides adjacent to the corner is now one shorter side. So instead of having two full sides adjacent to the corner, we have two truncated sides and the new cut side. Therefore, total sides: original four sides minus two sides (which were adjacent to the corner) plus three sides (the two truncated sides and the new cut side). Wait, 4 - 2 + 3 = 5. So the resulting shape is a pentagon. So in this case, n=4 becomes 5, which is n + 1.Another example: a pentagon (5 sides). Cut off one corner. The resulting shape would be a hexagon (6 sides). So n becomes n + 1. Wait, but the proposition says it could be either n - 1 or n + 1. So when would it be n - 1?Wait, maybe if the cut is made in such a way that it merges two sides? For example, if the cut is not just truncating a corner but actually connects two non-adjacent sides. But in a convex polygon, any straight line cutting off a corner would have to go from one side to an adjacent side, thereby increasing the number of sides by 1. However, in a concave polygon, perhaps cutting off a corner could result in a different outcome.Wait, consider a concave polygon. Let's say a star-shaped polygon. If you cut off a "corner" which is actually a reflex vertex (interior angle greater than 180 degrees), then cutting a line from one side to another might actually decrease the number of sides. For example, suppose you have a polygon with a "dent" in it, a concave vertex. Cutting off that dent with a straight line might remove two sides and replace them with one, decreasing the total number of sides by 1.Wait, let's think of a specific example. Imagine a polygon shaped like a house: a rectangle with a triangular roof. So that's a pentagon. If the "roof" part is a triangle, making the top vertex a convex angle. If we cut off that top vertex with a straight line from one side of the roof to the other, that would replace the two sides of the roof with a single horizontal line, turning the pentagon into a rectangle, which is a quadrilateral. So n=5 becomes n=4, which is n - 1. So in this case, cutting off a convex vertex (the peak of the roof) resulted in reducing the number of sides by 1. Wait, but how?Wait, in this case, the original pentagon has five sides: the base (rectangle's bottom side), two vertical sides, the two slanted sides of the roof, and the top peak. If we cut off the peak with a horizontal line, connecting the two top ends of the slanted sides, we replace the two slanted sides with a single horizontal side. Therefore, the number of sides decreases by 1. So in this case, cutting off a corner (the peak) resulted in n - 1 sides.But in the previous examples with convex polygons, cutting off a corner increased the number of sides. So depending on the shape of the polygon (convex or concave), cutting off a corner can either increase or decrease the number of sides. Therefore, proposition (1) states that cutting off a corner results in either n - 1 or n + 1 sides, which seems to be correct. Because depending on whether the original polygon had a convex or concave vertex being cut off, the number of sides can either increase or decrease by 1.But wait, the problem states "a corner of an n-sided paper piece". A corner is a vertex. If the polygon is convex, all its vertices are convex. If it's concave, some vertices are concave (reflex). So cutting off a convex corner (as in the square example) increases the number of sides. Cutting off a concave corner (as in the house example) decreases the number of sides. Therefore, proposition (1) is correct: cutting off a corner (vertex) can result in either n - 1 or n + 1 sides, depending on whether the vertex was concave or convex.But wait, the original problem states "n ≥ 4". The house example is n=5. If we have a concave quadrilateral (n=4), like a chevron shape. Suppose you have a quadrilateral with one reflex angle. Cutting off that reflex vertex (concave corner) with a straight line. Let's visualize. Original quadrilateral has four sides, with one reflex angle. Cutting off that reflex corner would connect the two adjacent sides with a straight line, effectively removing the reflex angle and replacing two sides with one. So the number of sides would decrease by 1, from 4 to 3. Wait, but the problem states n ≥ 4. If you have a quadrilateral (n=4), cutting off a concave corner would result in a triangle (n=3). But the proposition claims the remaining piece is (n - 1) or (n + 1). So 4 - 1 = 3 or 4 + 1 = 5. But in this case, it becomes 3, which is allowed. So if n ≥ 4, the result can be n - 1 or n + 1. However, in the case of convex polygons, cutting a corner increases the number of sides; in concave polygons, cutting a concave corner can decrease the number of sides.Therefore, proposition (1) is correct because depending on the type of corner (convex or concave), the number of sides can either increase or decrease by 1. Therefore, proposition (1) is correct.Wait, but hold on. If you have a convex polygon with all convex corners, cutting any corner will always result in n + 1 sides. If you have a concave polygon with at least one concave corner, cutting a concave corner can result in n - 1 sides. However, if you have a concave polygon and you cut a convex corner, would that result in n + 1 sides? For example, take the house-shaped pentagon (n=5), which has one concave angle (the inside of the roof). If we cut off a convex corner (like one of the bottom corners of the rectangle part), that would result in a hexagon (n=6). So cutting a convex corner in a concave polygon still increases the number of sides. Therefore, in a concave polygon, cutting a convex corner: n + 1; cutting a concave corner: n - 1. Therefore, proposition (1) is correct because depending on which corner you cut, the number of sides can go up or down.Therefore, proposition (1) is correct. So I think proposition (1) is correct.Moving on to proposition (2): If x^{-|x - 3| - 1} = 1, then x = 1 or 3.We need to solve the equation x^{-|x - 3| - 1} = 1. Let's recall that for a^b = 1, possible cases are:1. a = 1, and b is anything.2. a = -1, and b is even.3. b = 0, and a ≠ 0.So let's consider these cases.Case 1: x = 1. Then, the base is 1, so 1 raised to any power is 1. Therefore, x=1 is a solution.Case 2: x = -1. Wait, but the equation is x^{-|x - 3| - 1} = 1. If x = -1, then base is -1. Let's check if this works. So exponent is -| -1 - 3| -1 = -| -4 | -1 = -4 -1 = -5. So (-1)^{-5} = -1, which is not equal to 1. So x = -1 is not a solution.Alternatively, maybe the base can be -1 if the exponent is even. But in this case, exponent is -5, which is odd, so even if exponent was even, negative exponent would mean reciprocal. So (-1) to any even exponent is 1, but if exponent is negative even, it's still 1. So maybe x = -1 could be a solution if exponent is even. Let's check.Suppose x is such that x = -1 and exponent is even. So exponent is -|x - 3| -1. If x = -1, exponent is -| -1 - 3| -1 = -4 -1 = -5, which is odd. So (-1)^{-5} = -1 ≠ 1. Therefore, x = -1 is not a solution.Case 3: exponent is 0. So -|x - 3| -1 = 0. Solving: -|x - 3| -1 = 0 → -|x - 3| = 1 → |x - 3| = -1. Which has no solution because absolute value cannot be negative.Case 4: x = 0. Then, 0 raised to any positive exponent is 0, but here exponent is -|0 - 3| -1 = -3 -1 = -4. So 0^{-4} is undefined. So x=0 is excluded.Other possibilities? Maybe x = 3. Let's check x = 3. Then exponent is -|3 - 3| -1 = -0 -1 = -1. So 3^{-1} = 1/3 ≠ 1. Wait, but x=3. So x=3, base is 3, exponent is -1. So 3^{-1} = 1/3 ≠ 1. Therefore, x=3 is not a solution. Wait, but the proposition says x=1 or 3. But according to this, x=3 is not a solution.Wait, hold on. Maybe there's another case. If x ≠ 0, then x^{-|x - 3| -1} = 1. So x^k = 1 implies either x=1, x=-1 with k even, or k=0. As we saw, x=-1 is invalid, k=0 is impossible. So the only solution is x=1. So the equation x^{-|x -3| -1}=1 only has solution x=1, right? But the proposition says x=1 or 3. Wait, where does x=3 come into play?Wait, let's check x=3 again. If x=3, then exponent is -|3 - 3| -1 = -0 -1 = -1, so 3^{-1}=1/3 ≠1. Therefore, x=3 is not a solution. So why does the proposition mention x=3?Alternatively, maybe when the exponent is 0, but we saw that's impossible. Or perhaps x=3 is a solution for another reason.Wait, maybe the equation is x^{- |x - 3| -1} =1. Let's rewrite the equation as x^{ - (|x - 3| +1) } = 1. So x^{k} =1 where k = -(|x -3| +1). So unless k=0, which is not possible, or x=1, or x=-1 with k even. As we saw, x=1 is the only solution. But according to the proposition, x=3 is also a solution. But when x=3, it's 3^{-1} =1/3≠1. So proposition (2) is incorrect because x=3 is not a solution. Therefore, proposition (2) is wrong. Therefore, only x=1 is a solution. Therefore, the answer given in the proposition (x=1 or 3) is incorrect.Therefore, proposition (2) is incorrect.Moving on to proposition (3): Given the function y=(2k - 3)x^{k - 3} + 2/x is an inverse proportion function of x, then k= 3/2.An inverse proportion function is typically y = k/x, where y is inversely proportional to x. So the function should be of the form y = A/x. So let's check the given function: y = (2k - 3)x^{k - 3} + 2/x. For this to be an inverse proportion function, all terms must be inverse proportional, i.e., proportional to 1/x. Therefore, the exponents of x in each term must be -1.So first term: (2k - 3)x^{k - 3}. The exponent here is (k - 3). For this to be an inverse proportion term, the exponent must be -1. So set k - 3 = -1 → k = 2.Second term: 2/x is already 2x^{-1}, which is inverse proportional. So if k=2, then the first term becomes (2*2 -3)x^{2 -3} = (4 -3)x^{-1} = 1*x^{-1} = 1/x. Therefore, the entire function becomes y = 1/x + 2/x = 3/x, which is indeed an inverse proportion function.But the proposition states that k=3/2. Let's check that. If k=3/2, then the first term becomes (2*(3/2) -3)x^{(3/2)-3} = (3 -3)x^{-3/2} = 0*x^{-3/2} = 0. Therefore, the function reduces to y=0 + 2/x = 2/x, which is indeed an inverse proportion function. Wait, but if k=3/2, the first term becomes zero, so the function is still inverse proportional. So in this case, even if k=3/2, the function is inverse proportional. Therefore, the equation is true for both k=2 and k=3/2?Wait, but let's check the problem statement. It says "the function y=(2k -3)x^{k -3} + 2/x is an inverse proportion function of x". An inverse proportion function is y = A/x. So if we have y = (2k -3)x^{k -3} + 2/x, for this to be equal to A/x, all terms must be proportional to 1/x. Therefore, any term with x raised to a power other than -1 must be zero. So:First term: (2k -3)x^{k -3}. To have this term proportional to 1/x, either:1. The coefficient (2k -3) is zero, making the entire term zero, or2. The exponent (k -3) is -1, making the term proportional to 1/x.Therefore, either:Case 1: 2k -3 =0 → k=3/2. In this case, the first term is 0, and the function becomes y=0 +2/x=2/x, which is inverse proportional.Case 2: k -3 = -1 →k=2. Then, the first term is (2*2 -3)x^{-1}= (4-3)/x=1/x. Therefore, the function becomes y=1/x +2/x=3/x, which is inverse proportional.Therefore, there are two solutions: k=3/2 and k=2. But the proposition states that k=3/2. So is the proposition correct? The problem says "then k=3/2". However, there's another solution at k=2. Therefore, unless there's a constraint that the function must only have the term 2/x, but if k=3/2, the first term disappears, but if k=2, the first term is non-zero but still inverse proportional. However, the problem says "is an inverse proportion function", which is true for both k=3/2 and k=2. Therefore, the proposition is incomplete because it only mentions k=3/2. Hence, proposition (3) is incorrect.But wait, let's check again. Maybe the problem considers "inverse proportion function" as strictly y = A/x, so if you have y = (2k -3)/x + 2/x, which simplifies to y = (2k -3 +2)/x = (2k -1)/x. So if k=3/2, then 2*(3/2) -1 =3 -1=2, so y=2/x. If k=2, then 2*2 -1=3, so y=3/x. So both k=3/2 and k=2 make the function an inverse proportion function. Therefore, the equation has two solutions. But the problem states "then k= 3/2", ignoring k=2. Therefore, the proposition is incorrect because there are two solutions. Therefore, proposition (3) is incorrect.Wait, but if k=2, then the function is y= (4 -3)x^{-1} +2/x =1/x +2/x=3/x. So yes, inverse proportional. If k=3/2, the first term is 0, so y=0 +2/x=2/x. So both are valid. Therefore, unless there's a constraint in the problem that the function is exactly 2/x, but the problem says "is an inverse proportion function of x", which both 2/x and 3/x are. Therefore, the answer is not unique, so the proposition stating k=3/2 is only partially correct. Therefore, the proposition is incorrect because there are multiple solutions.Therefore, proposition (3) is incorrect.Moving on to proposition (4): Given the quadratic function y=ax² +bx +c, and a>0, a−b+c<0, then b²−4ac ≤0.So we have a quadratic function with a>0 (opens upwards). The condition a - b + c <0. We need to determine if the discriminant b² -4ac ≤0.First, recall that for a quadratic function, the discriminant determines the nature of the roots. If discriminant is positive, two real roots; zero, one real root; negative, no real roots. If the quadratic is always positive (since a>0), then it has no real roots, so discriminant ≤0. However, the problem gives a condition a - b + c <0. Let's see what this condition implies.Note that a - b + c is equivalent to evaluating the quadratic at x = -1: f(-1) = a(-1)^2 + b(-1) + c = a - b + c. So f(-1) <0. Since the parabola opens upwards (a>0), if it has a point below the x-axis (f(-1) <0), then it must cross the x-axis twice, meaning discriminant is positive. Therefore, the discriminant should be >0, contradicting the proposition's conclusion that discriminant ≤0.Wait, let's verify this. If a quadratic function with a>0 has f(-1) <0, then since it opens upwards, it must cross the x-axis at two points (one on each side of the vertex). Therefore, discriminant must be positive. Therefore, the conclusion b² -4ac ≤0 is incorrect. Therefore, proposition (4) is incorrect.But let's formalize this. Suppose f(x) = ax² +bx +c, a>0. If f(-1) = a - b + c <0, then the quadratic must take a negative value at x=-1. Since the parabola opens upwards, it must cross the x-axis somewhere, so there are two real roots, hence discriminant is positive. Therefore, b² -4ac >0, which contradicts the proposition's statement that discriminant ≤0. Therefore, proposition (4) is incorrect.Therefore, among the four propositions, only proposition (1) is correct, and the rest are incorrect. Therefore, the number of correct propositions is 1, which corresponds to option (B).Final Answerboxed{B}</think>Let's analyze each proposition step by step:1. Proposition (1): If a corner of an ( n )-sided paper piece is cut off with scissors (( n geq 4 )), the remaining paper piece is an ((n-1))-sided or ((n+1))-sided polygon. - Cutting off a corner (vertex) of a polygon can either increase or decrease the number of sides depending on whether the vertex is convex or concave. For convex polygons, cutting off a vertex increases the number of sides by 1. For concave polygons, cutting off a reflex vertex can decrease the number of sides by 1. Therefore, this proposition is correct.2. Proposition (2): If ( x^{-mid x-3 mid -1} = 1 ), then ( x = 1 ) or 3. - Solving ( x^{-mid x-3 mid -1} = 1 ), we find that ( x = 1 ) is a solution because ( 1 ) raised to any power is ( 1 ). However, ( x = 3 ) is not a solution because ( 3^{-1} = frac{1}{3} neq 1 ). Therefore, this proposition is incorrect.3. Proposition (3): Given the function ( y = (2k - 3)x^{k - 3} + frac{2}{x} ) is an inverse proportion function, then ( k = frac{3}{2} ). - For the function to be an inverse proportion function, all terms must be proportional to ( frac{1}{x} ). This can be achieved if ( 2k - 3 = 0 ) (leading to ( k = frac{3}{2} )) or ( k - 3 = -1 ) (leading to ( k = 2 )). Since there are two solutions, the proposition is incorrect.4. Proposition (4): Given the quadratic function ( y = ax^2 + bx + c ) with ( a > 0 ) and ( a - b + c < 0 ), then ( b^2 - 4ac leq 0 ). - The condition ( a - b + c < 0 ) implies ( f(-1) < 0 ). Since the parabola opens upwards (( a > 0 )), it must intersect the x-axis, resulting in a positive discriminant. Therefore, this proposition is incorrect.Among the four propositions, only proposition (1) is correct. The number of correct propositions is (boxed{B}).,<think>Okay, let's try to figure out which of these four propositions are correct. I'll go through each one step by step. Starting with proposition (1): If a corner of an n-sided paper piece is cut off with scissors (n ≥ 4 and the cutting line is a straight line), the remaining paper piece is an (n-1)-sided or (n+1)-sided polygon.Hmm, so when you cut off a corner of a polygon, you're essentially replacing one edge with two new edges. Wait, no, actually, if you cut off a corner with a straight line, you're removing a vertex and replacing it with a new edge. Let me visualize this. For example, if you have a square (4 sides), and you cut off one corner with a straight line. The original square has four sides. Cutting off a corner would replace two edges (the ones meeting at the corner) with one new edge. Wait, no. Wait, cutting off a corner would actually remove one vertex and replace two edges with one new edge. So, for each corner you cut off, you reduce the number of sides by one. So, starting with an n-sided polygon, cutting off a corner would result in (n - 1) sides. But the proposition says it could be either (n - 1) or (n + 1). That doesn't seem right. Wait, maybe if the cut goes through more than one corner? But the problem says "a corner" is cut off. If you cut off a single corner with a straight line, you should be replacing two sides with one, so decreasing the number of sides by one. But maybe if the cut is such that it intersects adjacent sides, adding two new edges? Wait, no. Let me think again. Imagine a polygon. Each corner is a vertex where two sides meet. If you make a straight cut that removes the vertex, you're effectively cutting between two adjacent sides. The original two sides are now each split into two, but since you remove the vertex, you actually replace the two original sides with one new side. So, for example, cutting a corner off a square (4 sides) would result in a pentagon (5 sides)? Wait, that contradicts my previous thought. Wait, no. Wait, if you have a square, and you cut off a corner, you end up with a pentagon. Let me confirm. Imagine a square: four sides. If you cut off one corner, you replace the 90-degree angle with a diagonal cut. Now, instead of two sides (each adjacent to the corner), you have one new side where the cut was made. So the original two sides are now each shorter, but connected by the new cut. So that's three sides: the two shortened sides and the new cut. Wait, no. Wait, the original square has four sides. If you cut off a corner, you remove one vertex, and replace two sides with one new side. So the total number of sides becomes 4 - 1 = 3? But that can't be. Wait, a square is four sides. If you cut off a corner, you have five sides. Because each original side adjacent to the corner is split into two parts, but actually, when you cut the corner, you are adding a new edge where the cut is. Wait, maybe I need to think of it as removing one vertex and adding two new edges? Wait, no. Let's take a specific example. Take a square. Each corner is a vertex connected by two edges. If you make a straight cut from one side to another, removing the corner. Let's say you have the square with vertices A, B, C, D. If you cut off corner B by making a straight line from a point on AB to a point on BC. Then, the original sides AB and BC are each split into two segments, but the corner B is replaced by the new cut. So instead of AB and BC, you have two shorter sides and the new cut. So the total number of sides becomes: original four sides minus two sides (AB and BC) plus three sides (the two shortened sides and the new cut). Wait, no. Wait, the original square has four sides: AB, BC, CD, DA. If you cut off corner B by making a straight line from a point E on AB to a point F on BC. Then, the new polygon would have vertices A, E, F, C, D. So sides AE, EF, FC, CD, DA. That's five sides. So cutting off one corner increases the number of sides by one. So starting with n sides, cutting off a corner gives n + 1 sides. Wait, that contradicts my initial thought. So maybe the proposition is saying it could be either n - 1 or n + 1. How?Wait, maybe depending on how you cut. If the cut is such that it connects two non-adjacent edges? But the problem says "a corner" is cut off, so the cutting line must be intersecting two adjacent sides. Therefore, cutting off a corner would add one side. So the number of sides becomes n + 1. So for the square example, cutting off a corner gives a pentagon. Therefore, the proposition (1) is saying that it can be either n - 1 or n + 1. But according to my example, it's n + 1. So maybe the proposition is incorrect? Wait, but maybe in some cases, if the cut is along a diagonal, removing two corners? Wait, but the problem states that "a corner" is cut off, so only one corner is removed. So if you remove one corner, replacing two sides with one new side, but in reality, when you make the cut, you are actually adding a side. Wait, confusion arises here. Let me think again.Original polygon: n sides. Each corner is a vertex where two sides meet. Cutting off a corner with a straight line. The straight line will intersect two adjacent sides. Each of those sides is split into two parts. So, the original two sides become two truncated sides, and the cut creates a new side. So total sides: original n sides minus 2 sides (which are split) plus 3 sides (the two truncated parts and the new cut). Wait, that doesn't make sense. Wait, no. Wait, when you cut a corner, you replace two original sides with a new side. Wait, original polygon has n sides. Each corner is a vertex between two sides. If you remove the vertex by cutting between the two adjacent sides, you replace those two sides with one new side. So the total number of sides becomes n - 2 + 1 = n - 1. Wait, but in the square example, cutting off a corner gives a pentagon (n + 1). So which is correct?Wait, perhaps my mistake is in how I visualize the cut. Let's take a square. If I cut off a corner, say between side AB and BC. If I make a straight cut from a point on AB to a point on BC, closer to the corner B. Then, the original sides AB and BC are each split into two parts. So instead of AB and BC, you have two sides: from A to the cut point on AB, then the cut line to the cut point on BC, then from there to C. So, in total, each original side is split, but the number of sides increases by one. Because instead of AB and BC, you have three sides: the remaining part of AB, the cut, and the remaining part of BC. So instead of two sides, you have three. Therefore, total sides: original n sides minus 2 (AB and BC) plus 3 (the split parts and the cut) = n - 2 + 3 = n + 1. Therefore, in the square case, n = 4, so becomes 5. So in general, cutting off a corner increases the number of sides by 1. Therefore, proposition (1) says it can be (n - 1) or (n + 1). But according to this, it should only be n + 1. So proposition (1) is incorrect? But maybe there are cases where it can be n - 1? How?Wait, suppose instead of cutting between two adjacent sides, you make a cut that goes from one side to another non-adjacent side. But the problem states "a corner" is cut off, which implies that the cut is removing a single corner, so the cut must intersect two adjacent sides. Therefore, in that case, as per the square example, it's n + 1. So unless the original polygon has a different configuration. Wait, maybe for polygons with more sides. Let me try a pentagon. If I have a pentagon (5 sides) and cut off a corner. Using the same logic, replacing two sides with three, so increasing the number of sides by 1, so 5 becomes 6. So in general, n becomes n + 1. Therefore, proposition (1) is wrong because it says it can be n - 1 or n + 1. But according to my reasoning, it's always n + 1. So proposition (1) is incorrect. Hmm. Wait, but maybe if the cut is along an existing edge? Wait, but cutting a corner with a straight line, you have to intersect two sides. So if you cut along an existing edge, that's not cutting off a corner. So maybe the answer depends on how the cut is made. But according to standard results in geometry, cutting off a corner (vertex truncation) usually increases the number of sides by 1. Therefore, maybe proposition (1) is incorrect because it claims it could be either n - 1 or n + 1, but in reality, it's always n + 1. Therefore, proposition (1) is wrong. So (1) is incorrect.Moving on to proposition (2): If x^{-|x - 3|} = 1, then x = 1 or 3.So we need to solve the equation x^{-|x - 3|} = 1. Let's recall that for a^b = 1, possible cases are:1. a = 1, and b is anything.2. a = -1, and b is even.3. b = 0, and a ≠ 0.So let's check these cases for x^{-|x - 3|} = 1.Case 1: x = 1. Then, the exponent is -|1 - 3| = -2. So 1^{-2} = 1. So that works.Case 2: x = -1. Wait, but x is in the base here. If x = -1, then (-1)^{-| -1 - 3| } = (-1)^{-4} = 1. So that's also 1. Wait, but the proposition says x = 1 or 3. So maybe x = -1 is also a solution?But the problem didn't mention x = -1. Let's check.Wait, but x is the base, so x can be positive or negative, but if x is negative, then x must be raised to an integer exponent to result in a real number. However, in the equation x^{-|x - 3|} = 1, x can be negative only if the exponent is an integer. So let's see.Case 1: x = 1. As above, works.Case 2: x = -1. Exponent: -| -1 - 3 | = -4. So (-1)^(-4) = 1. So this is a valid solution.Case 3: Exponent is 0. So -|x - 3| = 0 => |x - 3| = 0 => x = 3. Then, 3^0 = 1. So x = 3 is a solution.Case 4: x = 0. But x cannot be 0 because 0 raised to a negative exponent is undefined.So solutions are x = 1, x = 3, and x = -1. Therefore, proposition (2) is missing x = -1. So the answer is not only x = 1 or 3, so proposition (2) is incorrect.Wait, but maybe x has to be positive? The problem didn't specify, but since the original equation is x^{-|x - 3|}, if x is negative, then the exponent must be an integer for the expression to be real. Let's see:If x is negative, then -|x - 3| is a negative integer (since |x - 3| is a positive integer if x is integer. But in general, if x is any real number, then |x - 3| is a non-negative real number, so exponent is -|x - 3|, which is non-positive. If x is negative, then x^{-|x - 3|} is a real number only if the exponent is an integer. But unless |x - 3| is an integer, the exponent isn't necessarily an integer. So maybe x is restricted to integers? The problem didn't specify. If x is allowed to be any real number, then x = -1 is a solution, but so could other x where the exponent is an even integer. For example, x = -2. Then exponent is -| -2 - 3 | = -5, so (-2)^{-5} = -1/32 ≠ 1. So that's not a solution. x = -1: exponent -4, (-1)^-4 = 1. x = -3: exponent -6, (-3)^{-6} = 1/((-3)^6) = 1/729 ≠ 1. So only x = -1 works. But maybe x can be non-integer. Let's take x = 2. Exponent is -|2 - 3| = -1, so 2^{-1} = 0.5 ≠ 1. x = 4: exponent -1, 4^{-1} = 0.25 ≠ 1. So the only solutions are x =1, x=3, and x=-1. Therefore, proposition (2) is incorrect because it misses x=-1.Wait, but maybe there's a mistake here. Let's check x=-1 again. If x=-1, then the left-hand side is (-1)^{-| -1 - 3 | } = (-1)^{-4} = 1, yes. But the problem didn't specify that x has to be positive, so x=-1 is a valid solution. Therefore, proposition (2) is incorrect.Moving to proposition (3): Given the function y = (2k - 3)x^{k - 3} + 2/x is an inverse proportion function of x, then k = 3/2.An inverse proportion function is typically of the form y = k/x, where k is a constant. So the function should have terms that are inversely proportional to x. Let's analyze the given function: y = (2k - 3)x^{k - 3} + 2/x.For this to be an inverse proportion function, all terms must be inversely proportional, i.e., proportional to 1/x. Therefore, any term with x raised to a power other than -1 should not be present. So:First term: (2k - 3)x^{k - 3}Second term: 2/x = 2x^{-1}So, for the entire function to be inverse proportion, the first term must either be zero or also proportional to x^{-1}. If the first term is zero, then 2k - 3 = 0 => k = 3/2. Alternatively, if the exponent k - 3 = -1, then k - 3 = -1 => k = 2. Let's check both possibilities.Case 1: 2k - 3 = 0 => k = 3/2. Then the first term becomes 0, and the function is y = 0 + 2/x = 2/x, which is inverse proportion. So k = 3/2 is a solution.Case 2: k - 3 = -1 => k = 2. Then the first term becomes (2*2 - 3)x^{-1} = (4 - 3)x^{-1} = x^{-1}, so the function is y = x^{-1} + 2/x = (1 + 2)/x = 3/x, which is also inverse proportion. Therefore, k = 2 is another solution.Therefore, there are two solutions: k = 3/2 and k = 2. However, proposition (3) states that k = 3/2. So the proposition is missing the solution k = 2. Therefore, proposition (3) is incorrect.Wait, but maybe the problem states that the function is an inverse proportion function, which usually refers to a single term. However, even if it's a sum of inverse proportions, it's still an inverse proportion function. But technically, if it's a sum of two inverse proportions, it's still inversely proportional. For example, y = 3/x can be seen as a single inverse proportion, but y = 1/x + 2/x is also 3/x. However, in the case where k = 2, the function becomes 3/x. So both k = 3/2 and k = 2 result in inverse proportion functions, so the proposition (3) is incorrect because k is not uniquely 3/2. Therefore, proposition (3) is wrong.Finally, proposition (4): Given the quadratic function y = ax² + bx + c, and a > 0, a - b + c < 0, then b² - 4ac ≤ 0.So we have a quadratic with a > 0 (opens upwards) and a - b + c < 0. The claim is that the discriminant is less than or equal to zero, meaning the quadratic has at most one real root.Let's analyze this. The condition a - b + c < 0. Let's note that for a quadratic function y = ax² + bx + c, the value at x = -1 is y(-1) = a(-1)^2 + b(-1) + c = a - b + c. So the condition a - b + c < 0 is equivalent to y(-1) < 0.Given that the parabola opens upwards (a > 0), and that y(-1) < 0, what does this say about the discriminant?If the quadratic has two real roots, then the parabola crosses the x-axis twice. Since it opens upwards, it would go below the x-axis between the two roots. However, y(-1) < 0 means that the point x = -1 is below the x-axis. But if the discriminant is positive (two real roots), then depending on where the vertex is, x = -1 could be between the roots or outside. Let's think of an example.Suppose the quadratic has two real roots, say at x = 0 and x = 2. Then the parabola opens upwards, and y(-1) would be a(-1)^2 + b(-1) + c = a - b + c. If the roots are at 0 and 2, then the quadratic can be written as y = a(x)(x - 2). Let's compute a - b + c:Expanding: y = a(x² - 2x) = ax² - 2ax + 0. So a - b + c = a - (-2a) + 0 = a + 2a = 3a > 0, which contradicts the given condition a - b + c < 0. Therefore, in this case, with two real roots, the value at x = -1 is positive, so y(-1) > 0. But we need y(-1) < 0. Therefore, perhaps if the quadratic has two real roots, y(-1) is positive, so if y(-1) < 0, the quadratic cannot have two real roots. Wait, but let's try another example. Suppose the quadratic has roots at x = -2 and x = 1. Then y = a(x + 2)(x - 1) = a(x² + x - 2). Then y(-1) = a(1 -1 - 2) = a(-2) = -2a < 0. Since a > 0, this satisfies y(-1) < 0. However, discriminant is b² - 4ac. Let's compute coefficients:y = a x² + a x - 2a. So b = a, c = -2a. Then discriminant is (a)^2 - 4*a*(-2a) = a² + 8a² = 9a² > 0. So discriminant is positive, but y(-1) = -2a < 0. This contradicts proposition (4), which claims that discriminant ≤ 0. Therefore, proposition (4) is incorrect.Therefore, in this example, a - b + c < 0 (since y(-1) = -2a < 0) but discriminant is 9a² > 0. Therefore, proposition (4) is false.So, summarizing:Proposition (1): Incorrect (should be n + 1 only, not n - 1 or n + 1).Proposition (2): Incorrect (missing x = -1).Proposition (3): Incorrect (another solution k = 2 exists).Proposition (4): Incorrect (counterexample exists where discriminant is positive).Therefore, all four propositions are incorrect. So the number of correct propositions is 0. The answer is (A) 0.Final Answerboxed{A}</think>Given the four propositions, we analyze each one step by step:1. Proposition (1): If a corner of an ( n )-sided paper piece is cut off with scissors (( n geq 4 ) and the cutting line is a straight line), the remaining paper piece is an ( (n-1) )-sided or ( (n+1) )-sided polygon. - Cutting off a corner of a polygon with a straight line replaces two adjacent sides with one new side, increasing the number of sides by 1. Therefore, the remaining polygon should be ( (n+1) )-sided. The proposition incorrectly states it could be ( (n-1) )-sided or ( (n+1) )-sided. Hence, this proposition is incorrect.2. Proposition (2): If ( x^{-|x-3|} = 1 ), then ( x = 1 ) or 3. - Solving ( x^{-|x-3|} = 1 ), we find solutions ( x = 1 ), ( x = 3 ), and ( x = -1 ). The proposition misses ( x = -1 ). Hence, this proposition is incorrect.3. Proposition (3): Given the function ( y = (2k - 3)x^{k-3} + frac{2}{x} ) is an inverse proportion function of ( x ), then ( k = frac{3}{2} ). - For the function to be inverse proportion, the term ( (2k - 3)x^{k-3} ) must be zero or also inversely proportional. Solutions are ( k = frac{3}{2} ) (making the term zero) and ( k = 2 ) (making the term inversely proportional). The proposition misses ( k = 2 ). Hence, this proposition is incorrect.4. Proposition (4): Given the quadratic function ( y = ax^2 + bx + c ), and ( a > 0 ), ( a - b + c < 0 ), then ( b^2 - 4ac leq 0 ). - The condition ( a - b + c < 0 ) implies ( y(-1) < 0 ). A counterexample with ( a > 0 ), ( y(-1) < 0 ), and positive discriminant shows the proposition is incorrect.All propositions are incorrect. The number of correct propositions is (boxed{A}).]

🔑:5.A.(1)The remaining paper piece can still be an n-sided polygon. The cutting method is shown in Figure 3.(2)When x=-1, -|x-3|=-4, the original equation also holds.(3)When k-3=-1, i.e., k=2, y=frac{3}{x}, y is also an inverse proportion function of x.(4)The original parabola opens upwards. When x=-1, y=a- b+c0.

🧠:[<think>Okay, so I need to simplify this expression: (x + 1 - 3/(x - 1)) divided by (x² - 4x + 4)/(x - 1). Then, pick a value from 0, 1, 2 to substitute and evaluate. Hmm. Let me take this step by step. First, let me parse the expression. The numerator is (x + 1 - 3/(x - 1)) and the denominator is ((x² - 4x + 4)/(x - 1)). So, this is a complex fraction. To simplify, maybe I should combine the terms in the numerator first so that it's a single fraction, and then divide by the denominator fraction. Starting with the numerator: x + 1 - 3/(x - 1). To combine these terms, I need a common denominator. The terms x and 1 can be considered as (x + 1)/1, so the common denominator would be (x - 1). Let me rewrite each term with denominator (x - 1):x = x*(x - 1)/(x - 1) = (x² - x)/(x - 1)1 = 1*(x - 1)/(x - 1) = (x - 1)/(x - 1)So, combining these:(x² - x)/(x - 1) + (x - 1)/(x - 1) - 3/(x - 1) = [ (x² - x) + (x - 1) - 3 ] / (x - 1)Let me compute the numerator inside the brackets:x² - x + x - 1 - 3 = x² - 0x - 4 = x² - 4So, the numerator becomes (x² - 4)/(x - 1)Therefore, the entire expression is:[ (x² - 4)/(x - 1) ] ÷ [ (x² - 4x + 4)/(x - 1) ]Dividing by a fraction is the same as multiplying by its reciprocal, so:(x² - 4)/(x - 1) * (x - 1)/(x² - 4x + 4)I notice that (x - 1) cancels out in the numerator and denominator:(x² - 4)/1 * 1/(x² - 4x + 4) = (x² - 4)/(x² - 4x + 4)Now, let's factor both the numerator and the denominator to see if they can be simplified further.Numerator: x² - 4. That's a difference of squares, so it factors into (x - 2)(x + 2).Denominator: x² - 4x + 4. That's a perfect square trinomial; it factors into (x - 2)².So, substituting these factors in:[(x - 2)(x + 2)] / [(x - 2)²] Here, the (x - 2) in the numerator cancels with one (x - 2) in the denominator, leaving:(x + 2)/(x - 2)So, the simplified expression is (x + 2)/(x - 2). Now, the next part is to choose a suitable value from 0, 1, 2 to substitute and evaluate. Let's check each value first in the original expression to ensure we don't have any undefined expressions or division by zero.Original expression: (x + 1 - 3/(x - 1)) ÷ (x² - 4x + 4)/(x - 1)Check for restrictions:The denominators in the original expression are (x - 1) in the first part and (x - 1) in the denominator fraction. Also, the denominator of the simplified expression is (x - 2). So, x cannot be 1 or 2, because these would make the denominators zero.Given the options 0, 1, 2, we need to exclude 1 and 2. But wait, 1 is excluded because x - 1 is in the denominator of the original expression, and 2 is excluded because x - 2 is in the denominator of the simplified expression. So, the only valid value to substitute is 0. Because 1 and 2 would make the expression undefined.Therefore, substitute x = 0 into the simplified expression (x + 2)/(x - 2):(0 + 2)/(0 - 2) = 2/(-2) = -1Let me confirm by substituting x = 0 into the original expression to make sure the result is the same.Original expression:Numerator: 0 + 1 - 3/(0 - 1) = 1 - 3/(-1) = 1 + 3 = 4Denominator: (0² - 4*0 + 4)/(0 - 1) = (0 - 0 + 4)/(-1) = 4/(-1) = -4So, the entire expression is 4 / (-4) = -1, which matches the simplified expression's result. So, x = 0 is valid and gives -1.Therefore, the suitable value is 0, and evaluating gives -1.Final Answerboxed{-1}</think>To simplify the expression ((x + 1 - frac{3}{x - 1}) div frac{x^2 - 4x + 4}{x - 1}):1. Combine the terms in the numerator: [ x + 1 - frac{3}{x - 1} = frac{x^2 - x + x - 1 - 3}{x - 1} = frac{x^2 - 4}{x - 1} ]2. Rewrite the division as multiplication by the reciprocal: [ frac{x^2 - 4}{x - 1} div frac{x^2 - 4x + 4}{x - 1} = frac{x^2 - 4}{x - 1} times frac{x - 1}{x^2 - 4x + 4} ]3. Simplify by canceling common factors: [ frac{(x - 2)(x + 2)}{(x - 2)^2} = frac{x + 2}{x - 2} ]Next, substitute a suitable value from 0, 1, 2. The values 1 and 2 are excluded because they make the denominator zero. Substituting (x = 0):4. Evaluate the simplified expression at (x = 0): [ frac{0 + 2}{0 - 2} = frac{2}{-2} = -1 ]Verification with the original expression confirms the result is correct.[boxed{-1}],<think>Alright, let's see. I need to simplify this expression: (x + 1 - 3/(x - 1)) divided by (x² - 4x + 4)/(x - 1). Then choose a value from 0, 1, 2 to substitute and evaluate. Hmm, okay. Let me start by understanding each part of the expression step by step.First, the numerator of the main fraction is (x + 1 - 3/(x - 1)). That looks like a combination of polynomial terms and a rational expression. The denominator is (x² - 4x + 4)/(x - 1). Let me recall that dividing by a fraction is the same as multiplying by its reciprocal, so maybe I can rewrite the whole expression as the numerator multiplied by the reciprocal of the denominator. That might make it easier to simplify.So, let me write it out:[(x + 1 - 3/(x - 1))] ÷ [(x² - 4x + 4)/(x - 1)] = [(x + 1 - 3/(x - 1))] * [(x - 1)/(x² - 4x + 4)]Before I proceed, I need to check if there are any restrictions on x. The denominators here are (x - 1) and (x² - 4x + 4). Since division by zero is undefined, x can't be 1 (because x - 1 = 0) and x can't make x² - 4x + 4 = 0. Let me solve that quadratic equation to find the other restrictions.x² - 4x + 4 = 0. That factors into (x - 2)² = 0, so x = 2 is a double root. Therefore, x cannot be 2 either. So, the domain restrictions are x ≠ 1 and x ≠ 2. That's important to note, especially when substituting values later. So when the problem says to choose a value from 0, 1, 2, we have to remember that 1 and 2 are excluded. Therefore, only 0 is suitable. Wait, the problem says "choose a suitable value from 0, 1, 2 to substitute in and evaluate." So they must expect 0 to be the only valid choice here. Let me keep that in mind.But let's first focus on simplifying the expression. Let's handle the numerator first: x + 1 - 3/(x - 1). To combine these terms, I need a common denominator. The terms x and 1 can be considered as (x + 1)/1, so the common denominator with 3/(x - 1) would be (x - 1). Therefore, I can rewrite x + 1 as (x + 1)*(x - 1)/(x - 1). Let's do that.So:x + 1 - 3/(x - 1) = [(x + 1)(x - 1) - 3]/(x - 1)Wait, let me check that. If I have:x + 1 = (x + 1)*(x - 1)/(x - 1) + 1*(x - 1)/(x - 1)? Wait, no. Wait, x is equivalent to x*(x - 1)/(x - 1), but that complicates things. Maybe I should handle x + 1 as two separate terms. Let me think again.Wait, maybe combining x + 1 into a single term first. But x + 1 is already a binomial. So perhaps to combine x + 1 and -3/(x - 1), I need to express x + 1 with denominator (x - 1). Let's see:x + 1 = (x + 1)*(x - 1)/(x - 1) = [x*(x - 1) + 1*(x - 1)]/(x - 1) = [x² - x + x -1]/(x - 1) = [x² - 1]/(x - 1)Wait, that seems complicated, but actually, x + 1 can be written as [ (x + 1)(x - 1) ] / (x - 1) if we multiply numerator and denominator by (x - 1). However, that would change the value unless we also subtract 3/(x - 1) properly. Let's try:x + 1 - 3/(x - 1) = [ (x + 1)(x - 1) - 3 ] / (x - 1)Is that correct? Let me verify. If I take x + 1 and write it over 1, then subtract 3/(x - 1), the common denominator is (x - 1). So:(x + 1) * (x - 1)/(x - 1) - 3/(x - 1) = [ (x + 1)(x - 1) - 3 ] / (x - 1)Yes, that works. So expanding (x + 1)(x - 1):(x + 1)(x - 1) = x² - 1. So, substituting back:[ x² - 1 - 3 ] / (x - 1) = (x² - 4)/(x - 1)Therefore, the numerator simplifies to (x² - 4)/(x - 1). Wait, x² - 4 factors into (x - 2)(x + 2). So, numerator becomes (x - 2)(x + 2)/(x - 1).So now, the entire expression is:[(x - 2)(x + 2)/(x - 1)] * [(x - 1)/(x² - 4x + 4)]Because the original expression is numerator times reciprocal of denominator. Let's verify:Original expression after simplifying numerator: [(x² - 4)/(x - 1)] * [(x - 1)/(x² - 4x + 4)]Yes. Now, notice that (x² - 4x + 4) is a quadratic that factors as (x - 2)^2. Let me confirm:x² -4x +4 = (x - 2)^2, yes. So substituting that in:[(x - 2)(x + 2)/(x - 1)] * [(x - 1)/(x - 2)^2)]Now, we can cancel terms here. The (x - 1) in the denominator cancels with the (x - 1) in the numerator. Similarly, (x - 2) in the numerator cancels with one of the (x - 2) in the denominator. Let's see:After cancellation:(x + 2)/ (x - 2)So, the simplified expression is (x + 2)/(x - 2). Let me check that again.Original expression: [(x + 1 - 3/(x - 1))] ÷ [(x² - 4x + 4)/(x - 1)]After simplifying numerator: (x² - 4)/(x - 1)Divided by (x² -4x +4)/(x -1) becomes multiply by reciprocal:(x² - 4)/(x -1) * (x -1)/(x² -4x +4) = (x² -4) * (x -1)/( (x -1)(x² -4x +4) )Cancel (x -1):(x² -4)/(x² -4x +4)Factor numerator and denominator:Numerator: x² -4 = (x -2)(x +2)Denominator: x² -4x +4 = (x -2)^2So:[(x -2)(x +2)] / [(x -2)^2] = (x +2)/(x -2)Yes, perfect. So the simplified expression is (x + 2)/(x - 2). That's much simpler.Now, the next part is to choose a suitable value from 0, 1, 2 to substitute and evaluate. But we have to remember the domain restrictions. The original expression is undefined at x =1 and x=2. So substituting x=1 or x=2 would make the original expression undefined, but after simplifying to (x +2)/(x -2), x=1 is allowed? Wait, no. Wait, the simplified expression (x +2)/(x -2) is defined except when x=2. However, the original expression was undefined at x=1 and x=2. Therefore, even though the simplified expression is defined at x=1, we must still exclude x=1 because the original expression was undefined there. So, when substituting, we can only use values that are in the domain of the original expression. Since 0,1,2 are given, and 1 and 2 are excluded, only x=0 is suitable.Therefore, substitute x=0 into the simplified expression (x +2)/(x -2):(0 +2)/(0 -2) = 2/(-2) = -1Therefore, the value of the expression at x=0 is -1.Let me double-check by substituting x=0 into the original expression to ensure that the simplification is correct.Original expression: (0 +1 - 3/(0 -1)) ÷ (0² -4*0 +4)/(0 -1)Calculate numerator: 0 +1 - 3/(-1) = 1 + 3 = 4Denominator: (0 -0 +4)/( -1 ) = 4/(-1) = -4Therefore, 4 divided by -4 is -1. Which matches the simplified expression's result. So the simplification is correct.Therefore, substituting x=0 gives -1, which is the answer.Final Answerboxed{-1}</think>To simplify the expression ((x + 1 - frac{3}{x - 1}) div frac{x^2 - 4x + 4}{x - 1}):1. Simplify the numerator: [ x + 1 - frac{3}{x - 1} ] Combine the terms over a common denominator ((x - 1)): [ frac{(x + 1)(x - 1) - 3}{x - 1} = frac{x^2 - 1 - 3}{x - 1} = frac{x^2 - 4}{x - 1} = frac{(x - 2)(x + 2)}{x - 1} ]2. Simplify the denominator: [ frac{x^2 - 4x + 4}{x - 1} = frac{(x - 2)^2}{x - 1} ]3. Divide by the denominator: [ frac{(x - 2)(x + 2)}{x - 1} div frac{(x - 2)^2}{x - 1} = frac{(x - 2)(x + 2)}{x - 1} times frac{x - 1}{(x - 2)^2} = frac{(x + 2)}{(x - 2)} ]4. Substitute a suitable value: The simplified expression is (frac{x + 2}{x - 2}). Considering the domain restrictions (x neq 1) and (x neq 2), the suitable value from (0, 1, 2) is (0).5. Evaluate at (x = 0): [ frac{0 + 2}{0 - 2} = frac{2}{-2} = -1 ]The value of the expression at (x = 0) is (boxed{-1}).]

🔑:To simplify the given expression (x+1-frac{3}{x-1})÷frac{x^2-4x+4}{x-1}, we follow these steps:1. First, we rewrite the numerator and denominator to make simplification easier: [ left(x+1-frac{3}{x-1}right)÷frac{x^2-4x+4}{x-1} = left(frac{x(x-1)+1(x-1)-3}{x-1}right) cdot frac{x-1}{(x-2)^2} ] Simplifying the numerator of the first fraction: [ = frac{x^2-x+x-1-3}{x-1} cdot frac{x-1}{(x-2)^2} ] Combining like terms: [ = frac{x^2-4}{x-1} cdot frac{x-1}{(x-2)^2} ]2. Factor the numerator of the first fraction and simplify the expression: [ = frac{(x+2)(x-2)}{x-1} cdot frac{x-1}{(x-2)^2} ] Cancel out the common terms (x-1) in the numerator and denominator: [ = frac{x+2}{x-2} ]Given the condition that x neq 1 and x neq 2 to avoid division by zero, we proceed to evaluate the expression for a suitable value of x.3. Substituting x = 0 into the simplified expression: [ frac{x+2}{x-2} = frac{0+2}{0-2} = frac{2}{-2} = -1 ]Therefore, when x = 0, the original expression evaluates to boxed{-1}.

🧠:[<think>Okay, so I need to find a four-digit number that when divided by 131 leaves a remainder of 112, and when divided by 132 leaves a remainder of 98. Hmm, let me think how to approach this.First, let me translate these conditions into mathematical equations. If the number is N, then:1. N ≡ 112 mod 131. That means N = 131k + 112 for some integer k.2. N ≡ 98 mod 132. So, N = 132m + 98 for some integer m.Since both expressions equal N, we can set them equal to each other: 131k + 112 = 132m + 98.Let me rearrange this equation to group like terms. Subtract 98 from both sides: 131k + 112 - 98 = 132m. That simplifies to 131k + 14 = 132m.Hmm, so 131k + 14 is divisible by 132. Which can also be written as 131k ≡ -14 mod 132. Alternatively, since 131 is congruent to -1 mod 132 (because 132 - 131 = 1), then:-1 * k ≡ -14 mod 132Which simplifies to:-k ≡ -14 mod 132Multiply both sides by -1:k ≡ 14 mod 132So, k = 132t + 14 for some integer t.Now, substitute back into the original equation for N:N = 131k + 112 = 131*(132t + 14) + 112Let me compute that:131*132t = 17292t131*14 = 1834So, N = 17292t + 1834 + 112 = 17292t + 1946Therefore, the general solution for N is N = 17292t + 1946.But we need N to be a four-digit number. So, N must be between 1000 and 9999 inclusive.Let's find the possible values of t.First, compute the minimal N:When t = 0: N = 1946. That's a four-digit number. Wait, 1946 is a four-digit number. Hmm. But let's check if it meets the conditions.Wait, 1946 divided by 131. Let's compute 131*14 = 1834. Then 1946 - 1834 = 112. So yes, the remainder is 112. And 1946 divided by 132: 132*14 = 1848. 1946 - 1848 = 98. So remainder 98. Wait, so 1946 actually satisfies both conditions. But 1946 is a four-digit number. But the problem says "a four-digit number," so is 1946 acceptable? Let me check.Yes, 1946 is a four-digit number. So, perhaps that's the answer? But wait, maybe there are more solutions. Let's check for t = 1: N = 17292*1 + 1946 = 17292 + 1946 = 19238. Wait, 19238 is a five-digit number, which is too big. So, t=0 gives N=1946, which is four-digit, and t=1 gives a five-digit number. So, is 1946 the only four-digit solution? Wait, but wait a second, 1946 divided by 132: let me check again. 132*14 = 1848. 1946 - 1848 = 98. Correct. So remainder 98. So, 1946 seems to satisfy both conditions. Then why does the problem say "find a four-digit number"? Maybe there are multiple solutions, but the smallest one is 1946. But maybe there's another one in the four-digit range. Let me check.Wait, our general solution is N = 17292t + 1946. The modulus here is 17292, which is the least common multiple of 131 and 132? Wait, 131 and 132 are consecutive integers, so they are coprime. Therefore, LCM(131, 132) = 131*132 = 17292. So, the solutions repeat every 17292. Therefore, the next solution after 1946 would be 1946 + 17292 = 19238, which is five digits, as before. Therefore, the only four-digit solution is 1946. Wait, but 1946 is a four-digit number. So why does the problem ask to "find a four-digit number"? Maybe the question expects another number? Wait, perhaps I made a mistake in my calculations. Let me verify again.First, starting with N ≡ 112 mod 131 and N ≡ 98 mod 132.So, N = 131k + 112. Then, substitute into the second equation: 131k + 112 ≡ 98 mod 132.Subtract 112: 131k ≡ -14 mod 132.But 131 ≡ -1 mod 132, so -k ≡ -14 mod 132 => k ≡14 mod 132. Hence, k = 132t +14.Therefore, N = 131*(132t +14) +112 = 131*132t +131*14 +112. Calculate 131*14: 130*14=1820, 1*14=14, total 1820+14=1834. Then 1834 +112=1946. Therefore, N=17292t +1946. So when t=0, N=1946; t=1, N=1946+17292=19238. So indeed, only t=0 gives a four-digit number. Therefore, 1946 is the solution.But wait, 1946 is a four-digit number, right? 1-9-4-6. Yes, four digits. Then why does the problem ask to find a four-digit number? Maybe there's a mistake here? Wait, maybe I did a miscalculation.Wait, let's check with another approach. Let's use the Chinese Remainder Theorem. Since 131 and 132 are coprime, there exists a unique solution modulo 131*132=17292. So the solutions are N ≡ 1946 mod 17292. So the smallest positive solution is 1946. But 1946 is a four-digit number. Wait, 1946 is indeed a four-digit number, so that's the answer. Therefore, the problem might just be expecting 1946 as the answer.But let me check again the division to ensure there's no mistake.First, divide 1946 by 131:131*14 = 1834, as before. 1946 - 1834 = 112. So remainder 112. Correct.Then divide 1946 by 132:132*14 = 1848. 1946 - 1848 = 98. So remainder 98. Correct.So yes, 1946 satisfies both conditions and is a four-digit number. Hence, the answer is 1946.But just to ensure there's no other four-digit number, let's check if adding 17292 would result in another four-digit number. 1946 + 17292 = 19238, which is a five-digit number, so that's beyond. Therefore, 1946 is the only four-digit number satisfying those conditions.Therefore, the answer is 1946.Final Answerboxed{1946}</think>To find a four-digit number ( N ) that leaves a remainder of 112 when divided by 131 and a remainder of 98 when divided by 132, we start by translating these conditions into mathematical equations:1. ( N equiv 112 pmod{131} ) implies ( N = 131k + 112 ) for some integer ( k ).2. ( N equiv 98 pmod{132} ) implies ( N = 132m + 98 ) for some integer ( m ).Equating the two expressions for ( N ):[131k + 112 = 132m + 98]Rearranging terms, we get:[131k + 14 = 132m]This can be written as a congruence modulo 132:[131k equiv -14 pmod{132}]Since ( 131 equiv -1 pmod{132} ), we simplify to:[-k equiv -14 pmod{132} implies k equiv 14 pmod{132}]Thus, ( k = 132t + 14 ) for some integer ( t ). Substituting back into the expression for ( N ):[N = 131(132t + 14) + 112 = 17292t + 1834 + 112 = 17292t + 1946]We need ( N ) to be a four-digit number, so ( 1000 leq N leq 9999 ). Checking the smallest value when ( t = 0 ):[N = 1946]This is a four-digit number. Checking the next value when ( t = 1 ):[N = 17292 cdot 1 + 1946 = 19238]This is a five-digit number, which is too large. Therefore, the only four-digit solution is ( 1946 ).Verifying:- ( 1946 div 131 ) leaves a remainder of 112.- ( 1946 div 132 ) leaves a remainder of 98.Thus, the answer is (boxed{1946}).,<think>Okay, let's see. I need to find a four-digit number. So the number is between 1000 and 9999 inclusive. The problem says that when this number is divided by 131, the remainder is 112. And when divided by 132, the remainder is 98. Hmm, right. So I need to find such a number that satisfies both conditions. Let me try to write down the equations. If the number is N, then:N ≡ 112 mod 131andN ≡ 98 mod 132So, N can be written as 131k + 112 for some integer k. Similarly, N can also be written as 132m + 98 for some integer m. Since both expressions equal N, we can set them equal to each other:131k + 112 = 132m + 98Let me rearrange this equation:131k - 132m = 98 - 112131k - 132m = -14Hmm, so 131k - 132m = -14. I need to find integers k and m that satisfy this equation. Then, substituting back into N = 131k + 112 will give me the number. But since N has to be four digits, I need to also find the appropriate k such that N is between 1000 and 9999.First, let's try to solve the equation 131k - 132m = -14. This looks like a linear Diophantine equation. The general method for solving such equations is to use the extended Euclidean algorithm to find solutions. Let me recall how that works.First, we need to check if the greatest common divisor (gcd) of 131 and 132 divides -14. The gcd of 131 and 132 is 1, since they are consecutive integers and 131 is a prime number. Wait, 131 is a prime? Let me check. 131 divided by 2 is 65.5, not integer. Divided by 3: 131/3 ≈ 43.666. Divided by 5: ends with 5 or 0, so no. 7: 131/7 ≈ 18.714. 11: 131/11 ≈ 11.909. 13: 131/13 ≈ 10.077. So yes, 131 is a prime number. So the gcd(131,132)=1. Since 1 divides -14, there are solutions to this equation.So, to solve 131k - 132m = -14, we can write this as 131k + (-132)m = -14. The extended Euclidean algorithm can help find integers k and m that satisfy this. Let me try to apply the extended Euclidean algorithm to 131 and 132.Wait, 132 - 131 = 1. So, 132 - 131 = 1. Therefore, 1 = 132 - 131. Multiply both sides by -14: -14 = -14*132 + 14*131. So:-14 = 14*131 -14*132So comparing with 131k + (-132)m = -14, we can write this as 131*14 + (-132)*(-14) = -14. Wait, actually, let me check:If we have 14*131 -14*132 = -14. Let's compute:14*131 = 183414*132 = 18481834 - 1848 = -14. Yes, correct.Therefore, one solution is k = 14 and m = 14. So, k = 14, m =14 is a particular solution.But the general solution to the equation 131k -132m = -14 will be given by:Since the coefficients are 131 and -132, and the gcd is 1, the solutions will be periodic with period 132 for k and 131 for m. Wait, the general solution for equations of the form ax + by = c is given by x = x0 + (b/d)t and y = y0 - (a/d)t, where d is the gcd(a,b). Here, a = 131, b = -132, d =1. So, the general solution for k and m would be:k = k0 + (132/1)t = 14 + 132tm = m0 + (131/1)t =14 +131tWait, let me confirm. So, since our equation is 131k -132m = -14, which can be rewritten as 131k + (-132)m = -14. The standard form is ax + by = c, where a=131, b=-132, c=-14. Then, the general solution is:x = x0 + (b/d)t = 14 + (-132)/1 t =14 -132tWait, no. Wait, the standard formula is:If ax + by = c has a particular solution (x0, y0), then all solutions are given by:x = x0 + (b/d)ty = y0 - (a/d)tBut in our case, the equation is 131k -132m = -14. Let me write this as 131k + (-132)m = -14. So a=131, b=-132, c=-14. So then, the general solution is:k = k0 + ( (-132)/d ) tm = m0 + (131/d ) tBut since d=1, this becomes:k = k0 -132tm = m0 +131tBut our particular solution was k0=14, m0=14. So substituting in:k =14 -132tm=14 +131tTherefore, the general solution is k=14 -132t, m=14 +131t for any integer t.But since k and m must be integers, t can be any integer. Now, since N =131k +112, substituting k=14 -132t:N=131*(14 -132t)+112=131*14 -131*132t +112.Compute 131*14: 131*10=1310, 131*4=524, so 1310+524=1834. Then, 1834 -131*132t +112. Let's compute 131*132: 131*100=13100, 131*30=3930, 131*2=262, so 13100+3930=17030+262=17292. Therefore, N=1834 -17292t +112=1834+112 -17292t=1946 -17292t.Wait, so N=1946 -17292t. But since N is a four-digit number, 1000 ≤ N ≤9999. Let's find t such that N is within this range.So, 1000 ≤1946 -17292t ≤9999First, solve the left inequality: 1946 -17292t ≥10001946 -1000 ≥17292t946 ≥17292tt ≤946/17292 ≈0.0547. Since t must be an integer, t ≤0Similarly, the right inequality: 1946 -17292t ≤9999-17292t ≤9999 -1946=8053-17292t ≤8053Multiply both sides by -1 (remember to reverse the inequality):17292t ≥-8053t ≥-8053/17292≈-0.466. Since t must be an integer, t ≥0Wait, so from the inequalities, t must be ≤0 and ≥0. Hence, t=0.But if t=0, then N=1946 -0=1946. So, N=1946. Is 1946 a four-digit number? Yes. Let's check if this satisfies the original conditions.First, divide 1946 by 131. Let's compute 131*14=1834. 1946 -1834=112. So remainder 112. Good.Then, divide 1946 by 132. 132*14=1848. 1946 -1848=98. Remainder 98. Perfect. So 1946 satisfies both conditions.But wait, is 1946 the only solution? Because when t=0, it's 1946. If t=1, N=1946 -17292= -15346, which is negative. Not a four-digit number. If t=-1, N=1946 +17292=19238. Wait, 19238 is a five-digit number, which is beyond 9999? Wait, 19238 is actually five digits: 1-9-2-3-8. So 19238 is more than 9999. Wait, 9999 is the maximum four-digit number. So 19238 is larger, so invalid. Therefore, the only t that gives N in four digits is t=0, leading to N=1946.Wait, but 1946 is a four-digit number. However, the problem says "Find a four-digit number". Wait, is there another number? Because modulus equations can sometimes have multiple solutions within a range. Wait, but in this case, since 131 and 132 are consecutive integers, their least common multiple is 131*132=17292. So the solutions are spaced 17292 apart. Therefore, in the range of four-digit numbers, which is 1000 to 9999, the only solution would be 1946, and the next one would be 1946 +17292=19238, which is five digits. So 19238 is too big. Therefore, the only four-digit number is 1946.Wait, but let me double-check. Maybe I made a mistake in the general solution. Let's check.We had the equation 131k -132m = -14. We found a particular solution k=14, m=14. Then, since the homogeneous equation is 131k -132m=0, the solutions to this are k=132t, m=131t for integer t. Therefore, the general solution is k=14 +132t, m=14 +131t. Wait, hold on, maybe I made a mistake here earlier.Wait, let me recall. For the equation ax + by = c, once a particular solution (x0, y0) is found, the general solution is x = x0 + (b/d)t, y = y0 - (a/d)t where d is the gcd(a,b). So in our case, the equation is 131k -132m = -14. Let's write this as 131k + (-132)m = -14. So a=131, b=-132, d=1. Then, the general solution is:k = k0 + ( -132 / 1 ) t = k0 -132tm = m0 + (131 / 1 ) t = m0 +131tWhere (k0, m0) is a particular solution. We found (14,14) as a particular solution. Therefore, the general solution is:k=14 -132tm=14 +131tTherefore, substituting back into N=131k +112:N=131*(14 -132t) +112= 131*14 -131*132t +112= 1834 -17292t +112=1946 -17292tSo yes, the same result as before. So N=1946 -17292t. Therefore, as t increases, N decreases, and as t decreases (i.e., becomes negative), N increases.So, for t=0: N=1946t=1: N=1946 -17292= negativet=-1: N=1946 +17292=19238, which is five-digit.Therefore, only t=0 gives a four-digit number.Wait, but 1946 is a four-digit number. Let me confirm the division:1946 divided by 131: 131*14=1834, 1946-1834=112. Correct.1946 divided by 132: 132*14=1848, 1946-1848=98. Correct.So, 1946 is the number. But wait, the problem says "Find a four-digit number...". So the answer is 1946. But let me check again if there's another solution. Since the modulus are 131 and 132, which are coprime, the Chinese Remainder Theorem tells us there's a unique solution modulo 131*132=17292. Therefore, in the range from 0 to 17291, there's only one solution, which is 1946. Since the next solution would be 1946 +17292=19238, which is five-digit, so not allowed. Therefore, 1946 is the only four-digit number that satisfies the conditions.But wait, hold on. Let me check with another approach to make sure.Alternative approach: Let me express N as 131k +112. Since N ≡98 mod132, substitute N into the second congruence:131k +112 ≡98 mod132So, 131k ≡98 -112 mod132131k ≡-14 mod132131k ≡118 mod132 (since -14 +132=118)Now, 131 ≡-1 mod132, because 131=132-1. Therefore, 131k ≡-k mod132. Therefore:- k ≡118 mod132Multiply both sides by -1:k ≡-118 mod132-118 mod132 is the same as 132-118=14 mod132. Therefore, k≡14 mod132.Therefore, k=14 +132t for some integer t.Therefore, substituting back into N=131k +112:N=131*(14 +132t) +112=131*14 +131*132t +112=1834 +17292t +112=1946 +17292tWait, wait. Here, in this approach, we get N=1946 +17292t. Wait, this is different from the previous expression where N=1946 -17292t. Hmm, why the discrepancy?Wait, probably because in the first approach we had k=14 -132t leading to N=1946 -17292t. Here, k=14 +132t leading to N=1946 +17292t. So which one is correct?Wait, in the second approach, we solved the congruence and found k≡14 mod132. Therefore, k=14 +132t. Therefore, substituting into N=131k +112 gives N=1946 +17292t. Therefore, the general solution is N=1946 +17292t. Therefore, to find four-digit numbers, solve 1000 ≤1946 +17292t ≤9999.So, subtract 1946: 1000 -1946 ≤17292t ≤9999 -1946-946 ≤17292t ≤8053Divide by 17292:-946 /17292 ≈-0.0547 ≤ t ≤8053 /17292≈0.466Therefore, t must be 0, since t must be integer. Hence, N=1946 +0=1946. So, same result. So in this approach, the next solution would be t=1: 1946 +17292=19238, which is five-digit. So 1946 is the only four-digit solution.So both approaches lead to N=1946 as the only four-digit solution. Therefore, the answer is 1946.But let me check again. Maybe the first approach had a sign error? Let me re-examine.First approach:We started with 131k -132m = -14.Found particular solution k=14, m=14.Then, the general solution is k=14 -132t, m=14 +131t. Then, N=131k +112=1946 -17292t.Second approach via congruence: k ≡14 mod132, so k=14 +132t, N=1946 +17292t.So, depending on how t is defined, the solutions can be written as N=1946 +17292t or N=1946 -17292t. Since t can be any integer, positive or negative, both expressions are equivalent. For example, t=1 in the second expression gives N=1946 +17292(1)=19238, which is equivalent to t=-1 in the first expression: N=1946 -17292*(-1)=1946 +17292=19238. Therefore, both are correct; the difference is just the sign of t. Hence, when looking for four-digit numbers, both expressions give the same solution when t=0.Therefore, 1946 is indeed the only four-digit number satisfying both conditions. Therefore, the answer is 1946.But wait, let me check with actual numbers. Let's test N=1946:1946 ÷131: 131*14=1834, 1946-1834=112. Correct.1946 ÷132: 132*14=1848, 1946-1848=98. Correct.So yes, 1946 works. Any other four-digit number? Let's suppose someone thought that maybe adding the modulus would give another solution, but as we saw, adding 17292 (the LCM of 131 and 132) would exceed four digits. So 1946 is the only one.Therefore, the answer is 1946.Final AnswerThe four-digit number is boxed{1946}.</think>Given a four-digit number ( N ) that satisfies the following conditions:1. When divided by 131, the remainder is 112.2. When divided by 132, the remainder is 98.We start by expressing ( N ) in terms of these conditions:[N = 131k + 112][N = 132m + 98]Equating the two expressions, we get:[131k + 112 = 132m + 98]Rearranging gives:[131k - 132m = -14]This is a linear Diophantine equation. Using the extended Euclidean algorithm, we find a particular solution ( k = 14 ) and ( m = 14 ). The general solution for ( k ) and ( m ) is:[k = 14 - 132t][m = 14 + 131t]Substituting ( k ) back into the expression for ( N ):[N = 131(14 - 132t) + 112 = 1834 - 17292t + 112 = 1946 - 17292t]We need ( N ) to be a four-digit number (1000 ≤ ( N ) ≤ 9999). Solving for ( t ):[1000 leq 1946 - 17292t leq 9999]For ( t = 0 ):[N = 1946 - 17292 cdot 0 = 1946]Checking ( N = 1946 ):- When divided by 131: ( 1946 div 131 = 14 ) remainder 112.- When divided by 132: ( 1946 div 132 = 14 ) remainder 98.Both conditions are satisfied. The next possible solution would be ( 1946 + 17292 = 19238 ), which is a five-digit number and thus invalid.Thus, the four-digit number is (boxed{1946}).]

🔑:Let the required four-digit number be N, then N=131 cdot k+112=132 l+98 cdots (1), so l=frac{N-98}{132}130, which contradicts l<75. Therefore, k=l. Thus, k=l=14, N=131 times 14+112=1946.