$\dfrac{x^2}{1+\sqrt{2-x}}+\dfrac{(2-x)^2}{1+\sqrt{x}}=1$
$\left\{\begin{matrix}a=\sqrt{x} & \\ b=\sqrt{2-x} & \end{matrix}\right.a;b \ge 0$
$\rightarrow \left\{\begin{matrix}\dfrac{a^4}{b+1}+\dfrac{b^4}{a+1}=1 & \\
& \\ a^2+b^2=2 & \end{matrix}\right.$
$(a+b)^2 \le 2(a^2+b^2)=4 \rightarrow a+b \le 2$
$1=\dfrac{a^4}{b+1}+\dfrac{b^4}{a+1} \ge \dfrac{(a^2+b^2)^2}{a+b+2} \ge \dfrac{2^2}{2+2}=1$
dấu bằng xảy ra khi $a=b \Longleftrightarrow .... \Longleftrightarrow x=1$