Question

Giải PT :

\(\left\{{}\begin{matrix}\sqrt{x+y+1}+1=\left(x+y\right)^2+\sqrt{2\left(x+y\right)}\\x^2-xy=3\end{matrix}\right.\)

Answer 1

Đặt \(x+y=a\ge0\) ta được:

\(\sqrt{a+1}+1=a^2+\sqrt{2a}\)

\(\Leftrightarrow a^2-1+\sqrt{2a}-\sqrt{a+1}=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1\right)+\frac{a-1}{\sqrt{2a}+\sqrt{a+1}}=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+1+\frac{1}{\sqrt{2a}+\sqrt{a+1}}\right)=0\)

\(\Leftrightarrow a=1\Rightarrow x+y=1\Leftrightarrow y=1-x\)

Thay vào pt dưới:

\(x^2-x\left(1-x\right)=3\Leftrightarrow2x^2-x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=2\\x=\frac{3}{2}\Rightarrow y=-\frac{1}{2}\end{matrix}\right.\)