Square of sum: Solving the problem

Solve for x:

$(x+3)^2=x^2+9$

$x=0$

What is the value of x?

$(x+3)^2=x^2+15$

$x=1$

Solve for x:

$(x+2)^2=x^2+12$

$x=2$

$(x-1)^2-(x+2)^2=15$

$x=-3$

$x^2+10x=-25$

$x=-5$

$(x+1)^2=x^2+13$

$x=6$

$4x^2=12x-9$

$x=\frac{3}{2}$

Solve for y:

$y^2+4y+2=-2$

$y=-2$

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

$\sqrt{3}$

Simply the following expression:

$(x+\sqrt{x})^2$

$x\lbrack x+2\sqrt{x}+1\rbrack$

Solve for x:

$x^2+32x=-256$

$x=-16$

$(x+2)^2-12=x^2$

$x=2$

$(x+1)^2=x^2$

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

Look at the following equation:

$\frac{\sqrt{x}-\sqrt{x+1}}{x+1}=1$

This can also be written as:

$x[A(x+B)-x^3]=0$

Calculate A and B.

B=1 , A=4

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

Find X

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

Solve the following equation:

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

$-\frac{1}{2}[1\pm\sqrt{5}\rbrack$

Solve the following system of equations:

$\begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases}$

$x=\frac{\sqrt{61}}{2}-2.5$

$y=\frac{\sqrt{61}}{2}+2.5$

or

$x=\frac{\sqrt{61}}{2}+2.5$

$y=\frac{\sqrt{61}}{2}-2.5$

Look at the following equation:

$\frac{\sqrt{x}+\sqrt{x+1}}{x+1}=1$

The same equation can be presented as follows:

$x[A(x+B)-x^3]=0$

Calculate A and B.

B=1 , A=4

Solve the following equation:

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

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