Square of sum: Using fractions

Consider the following relationship between x and y:

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

Express the equation in the form of a reduced multiplication formula.

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

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

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

$\frac{A}{X}+\frac{BX}{2}=\frac{(2X+3)^2}{X}-C$

Calculate the values of A, B, and C so that the equation is satisfied.

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

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

What is the product of x and y?

$xy=1$

$\frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64}$

Find X

$x=1,-\frac{17}{7}$

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

Calculate the product of x and y.

$xy=-1$

Solve the following equation:

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

$x=\frac{1}{2}$