1.
đặt $z=x+1$
$x^{2}+2x=y^{2}$
$\Leftrightarrow (x+1)^{2}-y^{2}=1 $
$\Leftrightarrow z^{2}-y^{2}=1$
$\Leftrightarrow (z-y).(z+y)=1$
$(1)\Leftrightarrow \frac{1}{(z-y)^{3}}+\frac{1}{(z+y)^{3}}=2z$
$\Leftrightarrow (\frac{1}{z-y}+\frac{1}{z+y})(\frac{1}{(z-y)^{2}}-\frac{1}{(z-y)(z+y)}+\frac{1}{(z+y)^{2}})=2z$
$\Leftrightarrow \frac{2z}{(z-y).(z+y)}.(\frac{1}{(z-y)^{2}}-1+\frac{1}{(z+y)^{2}})=2z $
$\Leftrightarrow \begin{bmatrix} 2z=0 & \\ \frac{1}{(z-y)^{2}}-1+\frac{1}{(z+y)^{2}}=1 & \end{bmatrix} $
$*z=0 \Rightarrow y^{2}=-1 $ (vô lí)
$*\frac{1}{(z-y)^{2}}-1+\frac{1}{(z+y)^{2}}=1 $
$\Leftrightarrow \frac{1}{(z-y)^{2}}+\frac{1}{(z+y)^{2}}=2$
mà $(z-y)(z+y)=1$ suy ra ...