Question

Giải hệ pt:

\(\left\{{}\begin{matrix}4x^2+y^2+2x+y=2\left(1-2xy\right)\\8\sqrt{1-2x}+y^2-9=0\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(x\le\dfrac{1}{2}\)

\(4x^2+y^2+2x+y=2-4xy\)

\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+2x+y-2=0\)

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

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

Thế vào pt dưới:

\(\Rightarrow\left[{}\begin{matrix}8\sqrt{y}+y^2-9=0\\8\sqrt{y+3}+y^2-9=0\end{matrix}\right.\)

\(\Leftrightarrow...\)