Question

helpp me

Answer 1

hpt \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2+3x=4+3y\\2x+3y=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2+5x=16\\y=\dfrac{12-2x}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{12-2x}{3}\right)^2+5x=16\\y=\dfrac{12-2x}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\left(12-2x\right)^2}{9}+5x=16\\y=\dfrac{12-2x}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}144-120x+25x^2=144-45x\\y=\dfrac{12-2x}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}25x^2-75x=0\\y=\dfrac{12-2x}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25x\left(x-3\right)=0\\y=\dfrac{12-2x}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\y=\dfrac{12-2x}{3}\end{matrix}\right.\)

Với x= 0, ta có: y=4

Với x=3, ta có: y= 2

KL: Nếu x=0 thì y=4

Nếu x=3 thì y=2