Question

Giải hệ \(\left\{{}\begin{matrix}y^2-5\sqrt{x}+5=0\\\sqrt{x+2}=\sqrt{y^2+2y+3}-\frac{1}{5}y^2+y\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

\(\sqrt{x+2}=\sqrt{y^2+2y+3}-\left(\sqrt{x}-1\right)+y\)

\(\sqrt{x}+\sqrt{x+2}=\sqrt{\left(y+1\right)^2+2}+y+1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\y+1=b\end{matrix}\right.\) \(\Rightarrow a+\sqrt{a^2+2}=b+\sqrt{b^2+2}\)

\(\Leftrightarrow a-b+\frac{\left(a-b\right)\left(a+b\right)}{\sqrt{a^2+2}+\sqrt{b^2+2}}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\\sqrt{a^2+2}+\sqrt{b^2+2}+a+b=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}=y+1\)

Thay vào pt đầu:

\(y^2-5\left(y+1\right)+5=0\Rightarrow...\)