1) $\dfrac{1}{x+1}=\dfrac{1+x}{9}$ $ĐKXĐ : x \neq -1$
$⇒ (x+1)^2=9$
$⇔ \left[ \begin{array}{l}x+1=3\\x+1=-3\end{array} \right.$ $⇔ \left[ \begin{array}{l}x=2\\x=-4\end{array} \right.$ ( Thỏa mãn )
2)
Theo BĐT AM-GM ta có :
$\dfrac{x^2}{y} + y ≥ 2\sqrt[]{\dfrac{x^2}{y}.y} = 2x$
$\dfrac{y^2}{x}+x ≥ 2\sqrt[]{\dfrac{y^2}{x}.x}=2y$
$\to \dfrac{x^2}{y}+y+\dfrac{y^2}{x} + x≥ 2.(x+y)$
$\to \dfrac{x^2}{y}+\dfrac{y^2}{x} ≥ x+y$
Dấu "=" xảy ra $⇔x=y$