Question

Giải hệ \(\left\{{}\begin{matrix}\sqrt{x^2+2y+3}+2y-3=0\\2\left(2y^3+x^3\right)+3y\left(x+1\right)^2+6x\left(x+1\right)+2=0\end{matrix}\right.\)

Answer 1

\(2\left(2y^3+x^3\right)+3y\left(x+1\right)^2+6x\left(x+1\right)+2=0\)

\(\Leftrightarrow2\left(x^3+3x^2+3x+1\right)+3y\left(x+1\right)^2+4y^3=0\)

\(\Leftrightarrow2\left(x+1\right)^3+3\left(x+1\right)^2y+4y^3=0\)

Đặt \(x+1=a\)

\(\Rightarrow2a^3+3a^2y+4y^3=0\)

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

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

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

\(\Leftrightarrow x+1+2y=0\Rightarrow x=-2y-1\)

Thế vào pt đầu:

\(\sqrt{\left(-2y-1\right)^2+2y+3}=3-2y\)

\(\Leftrightarrow\sqrt{4y^2+6y+4}=3-2y\) (\(y\le\dfrac{3}{2}\))

\(\Leftrightarrow18y=5\)