point out the mistakes $1=2$? Q:prove $1=2$ ?

method1:-

let us consider $x=1$

then $x=x^2$

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

$x-1=(x-1)(x+1)$ 

$1=x+1$

finally 
$1=2$

i have little confusion here, is the mistake is considering $x=1$ or else any thing other than this.

method2:-

$-2=-2$

$1-3=4-6$

$(1^2)-(\frac{2*1*3}{2})=(2^2)-(\frac{2*2*3}{2})$

$(1^2)-(\frac{2*1*3}/2)+((\frac{3}{2})^2)=(2^2)-(\frac{2*2*3}{2})+((\frac{3}{2})^2)$

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

$1-\frac{3}{2}=2-\frac{3}{2}$ [here the formula if $x^n=y^n$ then $x=y$  is failed]

$1=2$

point out the mistakes in above two methods exactly and please explain
For the first: Don't divide by zero. (That's what you did going from $x-1=(x-1)(x+1)$ to $1=x+1$, since $x=1$.)

For the second: $x^2=y^2$ does not imply that $x=y$. After all, $(-1)^2=1=1^2,$ but $-1\ne 1$. The best we can conclude is that $x=\pm y$.
In the first one you divided by $x-1$, which is illegal cuz $x-1=0$.  I don't know what you're doing in the second one but I suspect you also divided by 0 there too lol.