Đáp án: $x=y=1$
Giải thích các bước giải:
Ta có:
$\sqrt{x+2}-\sqrt{y+1}=\sqrt{y+2}-\sqrt{x+1}$
$\rightarrow \sqrt{x+2}-\sqrt{y+2}=\sqrt{y+1}-\sqrt{x+1}$
$\rightarrow \dfrac{x+2-(y+2)}{\sqrt{x+2}+\sqrt{y+2}}=\dfrac{y+1-(x+1)}{\sqrt{y+1}+\sqrt{x+1}}$
$\rightarrow \dfrac{x-y}{\sqrt{x+2}+\sqrt{y+2}}=\dfrac{y-x}{\sqrt{y+1}+\sqrt{x+1}}$
$\rightarrow (x-y)(\dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}+\dfrac{1}{\sqrt{y+1}+\sqrt{x+1}})=0$
$\rightarrow x-y=0$
$\rightarrow x=y$
$\rightarrow \sqrt{x^2-(x+x)+5}+\sqrt{x-1}=2$
$\rightarrow \sqrt{x^2-2x+5}-2+\sqrt{x-1}=0$
$\rightarrow \dfrac{(\sqrt{x^2-2x+5})^2-2^2}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0$
$\rightarrow \dfrac{(x-1)^2}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0$
$\rightarrow \sqrt{x-1}(\dfrac{(x-1)\sqrt{x-1}}{\sqrt{x^2-2x+5}+2}+1=0)$
$\rightarrow x-1=0$
$\rightarrow x=1$
$\rightarrow x=y=1$
