\(\left(x-y\right)^2+2\cdot\frac{3}{2}\left(x-y\right)+\frac{9}{4}=4+\frac{9}{4}=\frac{25}{4}\)
\(\Rightarrow\left(x-y+\frac{3}{2}\right)^2=\frac{25}{4}\Rightarrow x-y+\frac{3}{2}=\frac{5}{2}\Rightarrow x-y=1\Rightarrow x=y+1\)
\(2x+3y=2\left(y+1\right)+3y=2y+2+3y=5y+2=12\Rightarrow5y=10\Rightarrow y=2\)
\(\Rightarrow x=y+1=2+1=3\)
vây x=23;y=2
Ta có : \(\hept{\begin{cases}\left(x-y\right)^2+3\left(x-y\right)=4\\2x+3y=12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^2+3\left(x-y\right)+\frac{9}{4}=4+\frac{9}{4}\\2x+3y=12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-y+\frac{3}{2}\right)^2=\frac{25}{4}\\2x+3y=12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+\frac{3}{2}=\frac{5}{2}\\2x+3y=12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y=\frac{5}{2}-\frac{3}{2}\\2x+3y=12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y=1\\2x+3y=12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-2y=2\\2x+3y=12\end{cases}}\)
<=> 2x - 2y - 2x - 3y = 2 - 12
<=> -5y = -10
<=> y = 2
=> 2x + 3.2 = 12
<=> 2.x + 6 = 12
<=> 2x = 6
<=> x = 3 .
