Sửa đề: \(\left\{{}\begin{matrix}2x+ay=1\\3ax-y=-2\end{matrix}\right.\)
Để hệ có nghiệm duy nhất thì \(\dfrac{2}{3a}\ne\dfrac{a}{-1}\)
=>\(3a^2\ne-2\)(luôn đúng)
=>Hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}2x+ay=1\\3ax-y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+ay=1\\3a^2\cdot x-a\cdot y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(3a^2+2\right)=-1\\3ax-y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{-1}{3a^2+2}\\y=3ax+2=\dfrac{-3a}{3a^2+2}+2=\dfrac{-3a+6a^2+4}{3a^2+2}=\dfrac{3a^2-6a+4}{3a^2+2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< 0\\y>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{3a^2+2}< 0\\\dfrac{3a^2-6a+4}{3a^2+2}>0\end{matrix}\right.\Leftrightarrow3a^2-6a+4>0\)
=>\(3a^2-6a+3+1>0\)
=>\(3\left(a^2-2a+1\right)+1>0\)
=>\(3\left(a-1\right)^2+1>0\)(luôn đúng)
=>\(a\in R\)
