Question

Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\\4\sqrt{2-x}+2\sqrt{2y+8}=\sqrt{9x^2+16}\end{matrix}\right.\)

Answer 1

ĐK:\(x\in\left[0;2\right];y\ge-4\)

\(pt\left(1\right)\Leftrightarrow\left(x-y-2\right)\left(x^2+xy+y^2-x+y+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}y=x-2\\\left(2x+y-1\right)^2-3\left(x-1\right)\left(x+y\right)\ge0\end{matrix}\right.\)

Thay \(y=x-2\) vào \(pt\left(2\right)\):

\(4\sqrt{2-x}+2\sqrt{2\left(x-2\right)+8}=\sqrt{9x^2+16}\)

\(\Rightarrow x=\dfrac{4\sqrt{2}}{3}\Rightarrow y=\dfrac{4\sqrt{2}}{3}-2\)