Square of Difference: Using fractions

$(\frac{1}{x}-x)^2=$

$\frac{x^4-2x^2+1}{x^2}$

Consider the following relationship between x and y:

$\frac{x}{4y}+\frac{y}{x}-1=0$

Choose the short multiplication formula that represents this equation.

$(x-2y)^2=0$

$\frac{\sqrt{2x^2-4xy+2y^2+(x-y)^2}}{x-y}=$

$\sqrt{3}$

$\frac{x^2+y^2}{(x-y)^2}=3,(x-y)^2=1$

What is the product of x and y?

$xy=1$

$\frac{A}{x}+\frac{Bx}{2}=\frac{(2x-3)^2}{x}-c$

Calculate the values of A, B, and C.

$A=9,B=8,C=-12$

$\frac{(\frac{1}{x}-\frac{1}{2})^2}{(\frac{1}{x}-\frac{1}{3})^2}=\frac{9}{4}$

Find X

2.5