a: Để hệ có nghiệm duy nhất thì \(\dfrac{a}{1}\ne\dfrac{2}{-a}\)
=>\(a^2\ne-2\)(luôn đúng)
Vậy: Hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}ax+2y=3\\x-ay=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=ay+4\\a\left(ay+4\right)+2y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=ay+4\\y\left(a^2+2\right)=3-4a\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{3-4a}{a^2+2}\\x=a\cdot\dfrac{3-4a}{a^2+2}+4=\dfrac{3a-4a^2+4a^2+8}{a^2+2}=\dfrac{3a+8}{a^2+2}\end{matrix}\right.\)
b: x>0 và y<0
=>\(\left\{{}\begin{matrix}\dfrac{3a+8}{a^2+2}>0\\\dfrac{3-4a}{a^2+2}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3a+8>0\\3-4a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a>-\dfrac{8}{3}\\4a>3\end{matrix}\right.\)
=>\(a>\dfrac{3}{4}\)
