answer:(1) The equation 3x^2 -32x -48=0 can be factored as: (x-12)(3x+4)=0. From this factorization, we get x-12=0 or 3x+4=0. Solving for x, we find x_1=12 and x_2 = -frac{4}{3}. So, the solutions are x=boxed{12} or x=boxed{-frac{4}{3}}.(2) The equation 4x^2 +x -3=0 can be factored as: (4x-3)(x+1)=0. From this factorization, we get 4x-3=0 or x+1=0. Solving for x, we find x_1=frac{3}{4} and x_2 = -1. So, the solutions are x=boxed{frac{3}{4}} or x=boxed{-1}.(3) The equation (3x+1)^2 -4=0 can be rewritten as: (3x+1)^2 = 4. Taking the square root of both sides, we have 3x+1=2 or 3x+1=-2. Solving for x, we find x_1=frac{1}{3} and x_2 = -1. So, the solutions are x=boxed{frac{1}{3}} or x=boxed{-1}.(4) The equation 9(x-2)^2 =4(x+1)^2 can be simplified by taking the square root of both sides: 3(x-2) = pm 2(x+1). This gives us two separate equations to solve: 3x-6=2x+2 and 3x-6=-2x-2. For the first equation 3x-6=2x+2, solving for x yields x=boxed{8}. For the second equation 3x-6=-2x-2, solving for x yields x=frac{4}{5}. So, the solutions are x=boxed{8} or x=boxed{frac{4}{5}}.

answer:Solution. If the numbers p, q, r are single-digit, then a=10 k+p, b=10 k+q, c=10 k+r, where 1 leq k leq 9. From a^{2}+b^{2}=c^{2} and p^{2}+q^{2}=r^{2}, it follows that 5 k+p+q=r. Squaring this equation, we get25 k^{2}+p^{2}+q^{2}+10 k p+10 k q+2 p q=r^{2}i.e.25 k^{2}+10 k p+10 k q+2 p q=0which is not possible. Therefore, p, q, r cannot be single-digit. We arrive at the same contradiction if we assume that p, q, r are two-digit.If p and q are single-digit and r is two-digit, then a=10 k+p, b=10 k+q,begin{gathered}c=100 k+r text {, for } 1 leq k leq 9 . text { From } a^{2}+b^{2}=c^{2} text { and } p^{2}+q^{2}=r^{2}, text { it follows that } 9800 k^{2}+200 r k=20 p k+20 q k,end{gathered}i.e.4900 k+100 r=p+qwhich is not possible, since p and q are single-digit numbers.If p is single-digit and q and r are two-digit, then a=10 k+p, b=100 k+q, c=100 k+r. From a^{2}+b^{2}=c^{2} and p^{2}+q^{2}=r^{2}, it follows that 5 k+p+10 q=10 r. Therefore, p must be divisible by 5, i.e., p=5, since p is single-digit. According to this, p^{2}=25=r^{2}-q^{2}=(r-q)(r+q). Since q, r geq 10, it is not possible that q+r=5 or q+r=1, which means q+r=25 and r-q=1, i.e., r=13, q=12.If q is single-digit and p and r are two-digit, then, as above, we get that q=5, r=13, p=12. Therefore, the required numbers are: p=5, q=12, r=13 or p=12, q=5, r=13.

answer:Answer: 25 (in the second variant 9). Note that z-t=(sqrt{} x-sqrt{} y)^{2} / 2. Denoting a=sqrt{} x, b=sqrt{} y, we get that a^{2}-b^{2}=3(a-b)^{2} / 2, which transforms into kappa(a-b)(2 a+2 b-3 a+3 b)=(a-b)(5 b-a)=0. By the condition a neq b, therefore a=5 b, i.e. x / y=25.

answer:A5. The rational function f(x)=frac{x^{2}-9}{x-1} has one pole at x=1, zeros at x=3 and x=-3. It intersects the y-axis at x=0. Since frac{x^{2}-9}{x-1}=frac{x^{2}-1-8}{x-1}=x+1-frac{8}{x-1}, y=x+1 is the oblique asymptote. The correct answer is (B).

answer:To solve the given problem, let's break down the solution step by step, closely following the standard solution provided:1. Simplify the given expression:Given expression: [left(x-2yright)left(x+2yright)-left(x-yright)^{2}+yleft(y+2xright)]div left(-2yright).First, expand the terms inside the brackets:begin{align*}&= [x^{2}-4y^{2}-(x^{2}-2xy+y^{2})+y^{2}+2xy]div left(-2yright) &= (x^{2}-4y^{2}-x^{2}+2xy-y^{2}+y^{2}+2xy)div left(-2yright) &= (4xy-4y^{2})div left(-2yright).end{align*}Simplify the expression further by dividing each term by -2y:begin{align*}&= 2y-2x.end{align*}2. Solve the equation left(x-2right)^{2}+|y-3|=0:Given that left(x-2right)^{2}+|y-3|=0, we can deduce that:begin{align*}x-2 &= 0, y-3 &= 0,end{align*}which implies:begin{align*}x &= 2, y &= 3.end{align*}3. Substitute x=2 and y=3 into the simplified expression:Substituting the values of x and y into the simplified expression 2y-2x, we get:begin{align*}&= 2times 3 - 2times 2 &= 6 - 4 &= 2.end{align*}Therefore, the final answer is boxed{2}.Analysis: We first simplified the given expression by expanding and combining like terms. Then, we solved the given equation to find the values of x and y. Finally, we substituted these values into the simplified expression to find the answer.Comments: This problem tests the ability to simplify and evaluate polynomials, requiring proficiency in the rules of addition, subtraction, multiplication, and division of polynomials. It is a basic type of question in polynomial manipulation.

answer:To solve for the product of the roots x_{1}x_{2} of the given quadratic equation x^{2}+x-1=0, we recall the formula for the product of the roots of a quadratic equation ax^{2}+bx+c=0, which is given by frac{c}{a}.Given that in our equation, a=1, b=1, and c=-1, we substitute these values into the formula to find the product of the roots:[x_{1}x_{2} = frac{c}{a} = frac{-1}{1} = -1.]Therefore, the product of the roots x_{1} and x_{2} is boxed{-1}.