Yes I can. But that's probably not what you are asking. So I'll just assume you wan't me to figure out the value x.
We have the following equality
[tex](\frac{x-1}{x+2})^2-\frac{3}{2}=\frac{x-1}{2x+4}[/tex].
Let's rearrange it so that we have terms with x on the left side and 3/2 on the right side
[tex](\frac{x-1}{x+2})^2-\frac{x-1}{2x-4}=\frac{3}{2}[/tex]
We now use this rule [tex](\frac{a}{b})^n=\frac{a^n}{b^n}[/tex] to epand the first fraction. In second fraction we just factor 2 out of the denominator
[tex]\frac{x^2-2x+1}{(x+2)^2}-\frac{x-1}{2(x+2)}=\frac{3}{2}[/tex].
Simplify further to get
[tex]\frac{2(x^2-2x+1)-((x+2)(x-1))}{(x+2)^2}=\frac{3}{2}\\\frac{2x^2-4x+2-(x^2+x-2)}{(x+2)^2}=\frac{3}{2}\\\frac{2x^2-4x+2-x^2-x+2}{(x+2)^2}=\frac{3}{2}\\\frac{x^2-5x+4}{(x+2)^2}=\frac{3}{2}[/tex]
Cross multiply to get
[tex]2(x^2-5x+4)=3((x+2)^2)[/tex]
Simplify even more
[tex]2x^2-10x+8=3x^2+12x+12\\\ -x^2-22x-4=0\\\ x^2+22x+4=0[/tex]
Now using quadratic formula we get two solutions
[tex]x_1 =\boxed{-11+3\sqrt{13}}, x_2=\boxed{-11-3\sqrt{13}}[/tex]
Hope this helps.
