💡Okay, so I have this problem about an ellipse, and I need to find the value of m. The equation of the ellipse is given as (x²)/25 + (y²)/(m²) = 1, and it's mentioned that m is greater than 0. The left focus of this ellipse is at the point F₁(-4, 0). Hmm, okay, let me try to remember what I know about ellipses.First, I recall that the standard form of an ellipse is (x²)/(a²) + (y²)/(b²) = 1, where a and b are the lengths of the semi-major and semi-minor axes, respectively. Now, depending on whether the major axis is along the x-axis or the y-axis, the positions of a and b switch. In this case, since the x² term is over 25 and the y² term is over m², I think that means the major axis is along the x-axis because 25 is larger than m² (assuming m is a positive real number). So, that would make a² = 25 and b² = m². Therefore, a = 5 and b = m.Next, I remember that for an ellipse, the distance from the center to each focus is given by c, where c² = a² - b². Since the foci are located along the major axis, which in this case is the x-axis, the foci are at (±c, 0). The problem gives the left focus at (-4, 0), so that means c = 4. Therefore, c² = 16.Now, plugging the values into the equation c² = a² - b², we have:16 = 25 - m²So, solving for m², I subtract 16 from both sides:m² = 25 - 16m² = 9Since m is greater than 0, taking the square root of both sides gives m = 3.Wait, let me double-check that. If a² is 25, then a is 5, and c is 4. So, c² is 16, which is less than a², which makes sense because in an ellipse, c is always less than a. Then, b² would be a² - c², which is 25 - 16 = 9, so b is 3. Since b is the semi-minor axis, and in the equation it's under y², that means the ellipse is wider along the x-axis, which matches the given focus points on the x-axis.So, yeah, m is 3. I think that's the answer.**Final Answer**The value of ( m ) is boxed{3}.

💡Okay, so I have this equation to simplify: [ a b^{2}left(frac{1}{(a+c)^{2}}+frac{1}{(a-c)^{2}}right)=a-b ]Alright, let's break this down step by step. First, I need to simplify the left-hand side of the equation. It looks like there are two fractions inside the parentheses that I can combine. So, the expression inside the parentheses is:[ frac{1}{(a+c)^{2}} + frac{1}{(a-c)^{2}} ]To add these two fractions, I need a common denominator. The denominators are ((a+c)^2) and ((a-c)^2), so the common denominator would be their product: ((a+c)^2(a-c)^2). Let me write that out:[ frac{1}{(a+c)^{2}} + frac{1}{(a-c)^{2}} = frac{(a-c)^2 + (a+c)^2}{(a+c)^2(a-c)^2} ]Now, I need to expand the numerators:First, expand ((a - c)^2):[ (a - c)^2 = a^2 - 2ac + c^2 ]Then, expand ((a + c)^2):[ (a + c)^2 = a^2 + 2ac + c^2 ]Adding these two together:[ (a^2 - 2ac + c^2) + (a^2 + 2ac + c^2) = 2a^2 + 2c^2 ]So, the combined fraction becomes:[ frac{2a^2 + 2c^2}{(a+c)^2(a-c)^2} ]I can factor out a 2 from the numerator:[ frac{2(a^2 + c^2)}{(a+c)^2(a-c)^2} ]Now, let's substitute this back into the original equation:[ a b^{2} cdot frac{2(a^2 + c^2)}{(a+c)^2(a-c)^2} = a - b ]Hmm, this looks a bit complicated. Maybe I can simplify the denominator. Notice that ((a + c)(a - c) = a^2 - c^2), so ((a + c)^2(a - c)^2 = (a^2 - c^2)^2). So, the equation becomes:[ a b^{2} cdot frac{2(a^2 + c^2)}{(a^2 - c^2)^2} = a - b ]Let me rewrite this for clarity:[ frac{2a b^{2} (a^2 + c^2)}{(a^2 - c^2)^2} = a - b ]Now, I need to solve for (b). This seems like a quadratic equation in terms of (b). Let's rearrange the equation to standard quadratic form.First, multiply both sides by ((a^2 - c^2)^2) to eliminate the denominator:[ 2a b^{2} (a^2 + c^2) = (a - b)(a^2 - c^2)^2 ]Expanding the right-hand side might be messy, but let me try:First, expand ((a - b)(a^2 - c^2)^2). Let me denote (D = a^2 - c^2) for simplicity.So, the right-hand side becomes:[ (a - b) D^2 = a D^2 - b D^2 ]Substituting back (D = a^2 - c^2):[ a(a^2 - c^2)^2 - b(a^2 - c^2)^2 ]So, the equation is:[ 2a b^{2} (a^2 + c^2) = a(a^2 - c^2)^2 - b(a^2 - c^2)^2 ]Let me bring all terms to one side to set the equation to zero:[ 2a b^{2} (a^2 + c^2) + b(a^2 - c^2)^2 - a(a^2 - c^2)^2 = 0 ]Now, this is a quadratic equation in terms of (b). Let me write it as:[ [2a(a^2 + c^2)] b^2 + [(a^2 - c^2)^2] b - [a(a^2 - c^2)^2] = 0 ]Let me denote:- ( A = 2a(a^2 + c^2) )- ( B = (a^2 - c^2)^2 )- ( C = -a(a^2 - c^2)^2 )So, the quadratic equation is:[ A b^2 + B b + C = 0 ]Using the quadratic formula:[ b = frac{-B pm sqrt{B^2 - 4AC}}{2A} ]Plugging in the values of A, B, and C:First, calculate the discriminant ( D = B^2 - 4AC ):[ D = [(a^2 - c^2)^2]^2 - 4 cdot 2a(a^2 + c^2) cdot (-a(a^2 - c^2)^2) ]Simplify term by term:First term: ([(a^2 - c^2)^2]^2 = (a^2 - c^2)^4)Second term: ( -4 cdot 2a(a^2 + c^2) cdot (-a(a^2 - c^2)^2) = 8a^2(a^2 + c^2)(a^2 - c^2)^2 )So, the discriminant becomes:[ D = (a^2 - c^2)^4 + 8a^2(a^2 + c^2)(a^2 - c^2)^2 ]Factor out ((a^2 - c^2)^2):[ D = (a^2 - c^2)^2 [ (a^2 - c^2)^2 + 8a^2(a^2 + c^2) ] ]Let me compute the expression inside the brackets:[ (a^2 - c^2)^2 + 8a^2(a^2 + c^2) ]First, expand ((a^2 - c^2)^2):[ a^4 - 2a^2c^2 + c^4 ]Then, expand (8a^2(a^2 + c^2)):[ 8a^4 + 8a^2c^2 ]Adding these together:[ a^4 - 2a^2c^2 + c^4 + 8a^4 + 8a^2c^2 = 9a^4 + 6a^2c^2 + c^4 ]Notice that (9a^4 + 6a^2c^2 + c^4) can be written as ((3a^2 + c^2)^2):[ (3a^2 + c^2)^2 = 9a^4 + 6a^2c^2 + c^4 ]So, the discriminant simplifies to:[ D = (a^2 - c^2)^2 (3a^2 + c^2)^2 ]Taking the square root of D:[ sqrt{D} = (a^2 - c^2)(3a^2 + c^2) ]Now, plug back into the quadratic formula:[ b = frac{ - (a^2 - c^2)^2 pm (a^2 - c^2)(3a^2 + c^2) }{ 2 cdot 2a(a^2 + c^2) } ]Simplify numerator:Factor out ((a^2 - c^2)):[ b = frac{ (a^2 - c^2) [ - (a^2 - c^2) pm (3a^2 + c^2) ] }{ 4a(a^2 + c^2) } ]Now, consider the two cases for the ± sign:1. **Case 1: Using the '+' sign**[ b = frac{ (a^2 - c^2) [ - (a^2 - c^2) + (3a^2 + c^2) ] }{ 4a(a^2 + c^2) } ]Simplify inside the brackets:[ -a^2 + c^2 + 3a^2 + c^2 = 2a^2 + 2c^2 = 2(a^2 + c^2) ]So,[ b = frac{ (a^2 - c^2) cdot 2(a^2 + c^2) }{ 4a(a^2 + c^2) } ]Cancel out (2(a^2 + c^2)):[ b = frac{ (a^2 - c^2) }{ 2a } ]2. **Case 2: Using the '-' sign**[ b = frac{ (a^2 - c^2) [ - (a^2 - c^2) - (3a^2 + c^2) ] }{ 4a(a^2 + c^2) } ]Simplify inside the brackets:[ -a^2 + c^2 - 3a^2 - c^2 = -4a^2 ]So,[ b = frac{ (a^2 - c^2) cdot (-4a^2) }{ 4a(a^2 + c^2) } ]Simplify:[ b = frac{ -4a^2(a^2 - c^2) }{ 4a(a^2 + c^2) } = frac{ -a(a^2 - c^2) }{ (a^2 + c^2) } ]Since (b > 0) is given, this solution would result in a negative value for (b) if (a > c), which contradicts the condition (b > 0). Therefore, we discard this solution.Thus, the only valid solution is:[ b = frac{a^2 - c^2}{2a} ]Which can be rewritten as:[ 2ab = a^2 - c^2 ]So, the simplified form of the given equation is:[ 2ab = a^2 - c^2 ]**Final Answer**The simplified expression is boxed{2ab = a^2 - c^2}.

💡Okay, so I have this problem about a grade with 4 liberal arts classes, each having 40 students. In each class, there are 8 boys and 32 girls. We're supposed to randomly sample 20 students from the entire grade, and then determine which of the options A, B, C, or D is correct.First, let me try to understand the setup. There are 4 classes, each with 40 students. So, in total, the grade has 4 classes × 40 students = 160 students. Each class has 8 boys and 32 girls, so overall, the entire grade has 4 classes × 8 boys = 32 boys and 4 classes × 32 girls = 128 girls. So, the grade is composed of 32 boys and 128 girls.Now, we're taking a random sample of 20 students from these 160. The question is about which of the statements A, B, C, or D is correct.Let me go through each option one by one.**Option A: At least one person from each class will be selected.**Hmm, so this is saying that in the sample of 20, we will have at least one student from each of the 4 classes. Is this necessarily true? Well, in probability terms, when we sample without replacement, it's possible, but not guaranteed, that we'll get at least one from each class. To check if it's definitely true, I need to see if it's possible to have a sample of 20 students that doesn't include at least one from each class. Since there are 4 classes, each with 40 students, is it possible to have a sample that only includes students from 3 classes? Let's see.The maximum number of students we could take from 3 classes is 3 × 40 = 120. But we're only taking 20 students. So, even if we took all 20 from one class, that's possible, right? But wait, each class only has 40 students, so 20 is less than 40. So, in fact, it's possible to have a sample of 20 students all from one class. Therefore, it's not guaranteed that we'll have at least one from each class. So, option A is not necessarily correct.**Option B: The number of girls selected will definitely be more than the number of boys.**So, this is saying that in the sample of 20, the number of girls will be more than the number of boys. Let's see.In the entire grade, there are 128 girls and 32 boys, so girls make up 80% of the student body, and boys make up 20%. When we take a random sample, we would expect the proportion to be similar, but it's not necessarily guaranteed.But the question is whether it's "definitely" more girls than boys. So, is it possible to have a sample where the number of boys is equal to or more than the number of girls?Let's calculate the maximum number of boys we could have in the sample. Since there are 32 boys in total, the maximum number of boys we could select is 32. But we're only selecting 20 students. So, the maximum number of boys in the sample is 20, but since there are only 32 boys, it's possible to have all 20 being boys? Wait, no, because there are only 32 boys in total, but we're only selecting 20. So, the maximum number of boys we could have is 20, but since 20 is less than 32, it's possible to have all 20 being boys. But wait, no, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys. But wait, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys. But wait, no, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys. But wait, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys.Wait, no, that's not correct. The total number of boys is 32, so if we're selecting 20 students, the maximum number of boys we could have is 20, but since 20 is less than 32, it's possible to have all 20 being boys. But wait, no, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys. But wait, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys.Wait, I'm getting confused. Let me think again. The total number of boys is 32, so the maximum number of boys we can select is 32, but since we're only selecting 20, the maximum number of boys in the sample is 20. But since 20 is less than 32, it's possible to have all 20 being boys. But wait, no, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys. But wait, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys.Wait, no, that's not correct. The total number of boys is 32, so if we're selecting 20 students, the maximum number of boys we could have is 20, but since 20 is less than 32, it's possible to have all 20 being boys. But wait, no, that's not correct because the total number of boys is 32, so 20 is less than 32, so it's possible to have all 20 being boys.Wait, I'm going in circles. Let me approach it differently. The probability of selecting a boy is 32/160 = 1/5, and a girl is 128/160 = 4/5. When sampling without replacement, the expected number of boys in the sample is 20 × (32/160) = 4. So, on average, we expect 4 boys and 16 girls. But the question is whether the number of girls will definitely be more than the number of boys. So, is it possible to have 10 or more boys in the sample?Let's calculate the maximum number of boys we could have. Since there are 32 boys, the maximum number of boys in the sample is 20, but that's only if all 20 are boys. But the probability of that is very low, but it's possible. However, the question is whether it's "definitely" more girls than boys. So, if it's possible to have 10 or more boys, then the number of girls wouldn't necessarily be more than the number of boys.Wait, but 20 students, if 10 are boys and 10 are girls, then girls are not more than boys. So, is it possible to have 10 boys in the sample? Let's see. The number of ways to choose 10 boys out of 32 and 10 girls out of 128 is C(32,10) × C(128,10). The total number of ways to choose 20 students is C(160,20). So, the probability is non-zero, meaning it's possible. Therefore, it's not guaranteed that the number of girls will be more than the number of boys. So, option B is not necessarily correct.**Option C: Given that Xiaowen is a boy and Xiaomei is a girl, the probability of Xiaowen being selected is less than that of Xiaomei being selected.**So, Xiaowen is a boy, and Xiaomei is a girl. We need to compare their probabilities of being selected in the sample.In a simple random sample without replacement, each individual has an equal probability of being selected. So, the probability of any specific student being selected is the same, regardless of gender. So, both Xiaowen and Xiaomei have the same probability of being selected, which is 20/160 = 1/8.Wait, but the option says that Xiaowen's probability is less than Xiaomei's. But since both have the same probability, this is not correct. Therefore, option C is incorrect.**Option D: If student A and student B are in the same class, and student C is in a different class, then the probability of both A and B being selected is the same as the probability of both A and C being selected.**Hmm, so we have three students: A and B are in the same class, and C is in a different class. We need to compare the probability that both A and B are selected versus the probability that both A and C are selected.Let me think about this. In a simple random sample, the probability of selecting any two specific students depends on whether they are in the same class or not.Wait, no, actually, in a simple random sample without replacement, the probability of selecting any two specific students is the same, regardless of their class. Because each pair of students has the same chance of being selected, regardless of their class.But wait, is that true? Let me think again. If A and B are in the same class, does that affect their joint probability of being selected compared to A and C being in different classes?Wait, no, because the sampling is done without regard to classes. Each student is equally likely to be selected, and the selection of one doesn't affect the selection of another, except for the fact that it's without replacement.Wait, but actually, the probability of selecting both A and B is C(158,18)/C(160,20), because we need to choose both A and B and then 18 more students from the remaining 158. Similarly, the probability of selecting both A and C is also C(158,18)/C(160,20). So, regardless of whether B and C are in the same class or not, the probability of selecting both A and B or both A and C is the same.Wait, but that doesn't make sense because if B and C are in the same class, their joint probability might be different? No, because the sampling is done across all classes. So, the fact that B and C are in the same class doesn't affect their individual probabilities of being selected with A.Wait, maybe I'm overcomplicating. Let me calculate it.The probability that both A and B are selected is:Number of ways to choose both A and B and 18 others: C(158,18)Total number of ways to choose 20 students: C(160,20)So, probability = C(158,18)/C(160,20)Similarly, the probability that both A and C are selected is also C(158,18)/C(160,20)Therefore, the probabilities are the same, regardless of whether B and C are in the same class or not.Wait, but that seems counterintuitive. If B and C are in the same class, does that affect their joint probability? Hmm, no, because the sampling is done without regard to classes. Each student is equally likely to be selected, regardless of their class. So, the fact that B and C are in the same class doesn't influence their joint probability with A.Therefore, option D is correct.Wait, but let me double-check. Suppose A and B are in the same class, and C is in a different class. Is the probability of selecting both A and B the same as selecting both A and C?Yes, because the selection is purely based on individual probabilities, and the classes don't influence the selection process. So, the joint probability of selecting any two specific students is the same, regardless of their class.Therefore, option D is correct.

💡First, I need to determine the total number of possible outcomes for the passenger's three rides. Since there are 4 cars and the passenger chooses one car each time independently, the total number of possible outcomes is (4 times 4 times 4 = 64).Next, I'll calculate the number of favorable outcomes where the passenger rides in 3 different cars. For the first ride, there are 4 possible cars. For the second ride, to ensure a different car, there are 3 remaining choices. For the third ride, there are 2 remaining choices to ensure it's different from the first two. Therefore, the number of favorable outcomes is (4 times 3 times 2 = 24).Finally, the probability that the passenger will ride in 3 different cars is the ratio of favorable outcomes to total outcomes, which is (frac{24}{64} = frac{3}{8}).

💡First, I need to determine the number of passengers on the bus after the changes at Jialejia station.Initially, there are ( m ) passengers on the bus. At the station, 12 passengers get off, so the number of passengers decreases by 12. This can be represented as ( m - 12 ).After some passengers get off, ( n ) new passengers board the bus. Therefore, the number of passengers increases by ( n ). This can be represented as ( m - 12 + n ).For the first part of the problem, the expression for the number of passengers on the bus now is ( m - 12 + n ).For the second part, I need to calculate the number of passengers when ( m = 26 ) and ( n = 6 ). Substituting these values into the expression:[m - 12 + n = 26 - 12 + 6 = 14 + 6 = 20]So, there are 20 passengers on the bus now.

💡To determine the coefficient of the ( x^4 ) term in the expansion of ( (1+x)^4 + (1+x)^5 ), I will use the binomial theorem.First, I'll find the coefficient of ( x^4 ) in ( (1+x)^4 ). According to the binomial theorem, the coefficient is given by ( binom{4}{4} ), which equals 1.Next, I'll find the coefficient of ( x^4 ) in ( (1+x)^5 ). Using the binomial theorem again, the coefficient is ( binom{5}{4} ), which equals 5.Finally, I'll add these two coefficients together: ( 1 + 5 = 6 ).Therefore, the coefficient of the ( x^4 ) term in the expansion is 6.