Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}4\sqrt{x+2}+2\sqrt{3\left(x+4\right)}=3y\left(y-1\right)+10\\\left(x+2\right)^3+x=y\left(y^2+1\right)-2\end{matrix}\right.\)

Answer 1

ĐKXĐ:...

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

\(\Leftrightarrow\left(x+2-y\right)\left[\left(x+2\right)^2+y\left(x+2\right)+y^2+1\right]=0\)

\(\Leftrightarrow\left(x+2-y\right)\left[\left(x+2+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\right]=0\)

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

Thay vào pt đầu:

\(4\sqrt{x+2}+2\sqrt{3\left(x+4\right)}=3\left(x+2\right)\left(x+1\right)+10\)

\(\Leftrightarrow3x^2+9x+16-4\sqrt{x+2}-2\sqrt{3\left(x+4\right)}=0\)

\(\Leftrightarrow3\left(x^2+2x+1\right)+2\left(x+3-2\sqrt{x+2}\right)+\left(x+7-2\sqrt{3x+12}\right)=0\)

\(\Leftrightarrow3\left(x+1\right)^2+\frac{2\left(x+1\right)^2}{x+3+2\sqrt{x+2}}+\frac{\left(x+1\right)^2}{x+7+2\sqrt{3x+12}}=0\)

\(\Rightarrow x=-1\)