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Question

$\begin{cases}(y-x)(xy+y+x)=2x+xy+1 \\ \sqrt{x} + \sqrt{y} -\sqrt{x^{2}+x} = 1 \end{cases}$

score 2 · Answer 1

Pt $(1)\Leftrightarrow y^2x-x+y^2-1-xy(x+1)-x(x+1)=0$

$\Leftrightarrow (x+1)(y^2-1)-xy(x+1)-x(x+1)=0$

$\Leftrightarrow (x+1)(y^2-1-xy-x)=0$

$\Rightarrow (y-1)(y+1)-x(y+1)=0$

$\Leftrightarrow (y+1)(y-1-x)=0$

$\Leftrightarrow y=x+1$

Thế vào $(2)\Leftrightarrow \sqrt{x}+\sqrt{x+1}-\sqrt{x(x+1)}=1$

$\Leftrightarrow \sqrt{x}-1-\sqrt{x+1}(\sqrt{x}-1)=0$

$\Leftrightarrow (\sqrt{x}-1)(1-\sqrt{x+1})=0$

1 Đáp án