Question

\(\left\{{}\begin{matrix}x^2y+y^3=x^4+x^6\\\left(x+2\right)\sqrt{y+1}=\left(x+1\right)^2\end{matrix}\right.\)

Answer 1

\(x^6-y^3+x^4-x^2y=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}y=x^2\\x=y=0\left(ktm\right)\end{matrix}\right.\)

Thay xuống dưới:

\(\left(x+2\right)\sqrt{x^2+1}=x^2+2x+1\)

Đặt \(x^2+1=t>0\Rightarrow t^2-\left(x+2\right)t+2x=0\)

\(\Leftrightarrow t\left(t-2\right)-x\left(t-2\right)=0\)

\(\Leftrightarrow\left(t-x\right)\left(t-2\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\left(vn\right)\\\sqrt{x^2+1}=2\Rightarrow...\end{matrix}\right.\)