\(2x^2+3xy+y^2=0\)
\(\Rightarrow2x^2+2xy+xy+y^2=0\)
\(\Rightarrow2x\left(x+y\right)+y\left(x+y\right)=0\)
\(\Rightarrow\left(x+y\right)\left(2x+y\right)=0\)
\(2x^2+3xy+y^2=0\)
\(\Leftrightarrow x^2+x^2+2xy+xy+y^2=0\)
\(\Leftrightarrow\left(x^2+xy\right)+\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow x\left(x+y\right)+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left(x+y\right)\left(2x+y\right)=0\)
Hoặc \(x+y=0\Leftrightarrow x=-y\left(1\right)\)
Hoặc \(2x+y=0\left(2\right)\)
Thế (1) vào (2) ta có:
\(-2y+y=0\)
\(\Leftrightarrow-y=0\Leftrightarrow y=0\)
\(\Leftrightarrow x=0\left(\text{vì x = -y}\right)\)
Vậy \(x=y=0\)
Ta có : \(2x^2+3xy+y^2=2x^2+2xy+xy+y^2=2x\left(x+y\right)+y\left(x+y\right)=\left(2x+y\right)\left(x+y\right)=0\)
\(=>\orbr{\begin{cases}2x+y=0\\x+y=0\end{cases}=>\orbr{\begin{cases}x=-\frac{y}{2}\\x=-y\end{cases}}}\)
Vậy x=-y hoặc x=-y/2 với mọi x thì 2x^2+3xy+y^2
