Question

Giải hệ phương trình :\(\left\{{}\begin{matrix}x^3-2x^2+2x+2y+x^2y-4=0\\x^2-xy-4x-1=\sqrt{3x-y+7}\end{matrix}\right.\)

Answer 1

\(x^3+2x-2x^2-4+x^2y+2y=0\)

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

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

\(\Leftrightarrow x+y-2=0\Rightarrow y=2-x\)

Thay vào pt dưới:

\(x^2-x\left(2-x\right)-4x-1=\sqrt{4x+5}\) (ĐKXĐ:...)

\(\Leftrightarrow2x^2-6x-1=\sqrt{4x+5}\)

\(\Rightarrow\left(2x^2-6x-1\right)^2=4x+5\)

\(\Leftrightarrow x^4-4x^3+3x^2+x-1=0\)

\(\Leftrightarrow\left(x^2-4x+1\right)\left(x^2-2x-1\right)=0\)