Đáp án: $\begin{cases}x=3\\y=-5\end{cases}$
Giải thích các bước giải:
$\sqrt{x-2}+\sqrt{4-x}+2\left( \sqrt{2x+y}+2\sqrt{x+1} \right)=3x+y+8$
Có: $\sqrt{x-2}+\sqrt{4-x}\le \sqrt{\left( {{1}^{2}}+{{1}^{2}} \right)\left( x-2+4-x \right)}$
$\Leftrightarrow \sqrt{x-2}+\sqrt{4-x}\le 2$
Có: $2\left( \sqrt{2x+y}+2\sqrt{x+1} \right)$$=2\left[ \sqrt{2x+y}+\sqrt{4\left( x+1 \right)} \right]$
$\Leftrightarrow 2\left( \sqrt{2x+y}+2\sqrt{x+1} \right)\le 2\left( \dfrac{1+2x+y}{2}+\dfrac{4+x+1}{2} \right)$
$\Leftrightarrow 2\left( \sqrt{2x+y}+2\sqrt{x+1} \right)\le 3x+y+6$
Vậy $VT\le 3x+y+8$
$\Leftrightarrow VT\le VP$
Dấu “=” xảy ra khi $\begin{cases}x-2=4-x\\2x+y=1\\x+1=4\end{cases}\Leftrightarrow\begin{cases}x=3\\y=-5\end{cases}$
