Để hệ có nghiệm duy nhất thì \(\dfrac{1}{a}\ne\dfrac{a}{-2}\)
=>\(a^2\ne-2\)(luôn đúng)
=>Hệ Phương trình luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}x+ay=2\\ax-2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-ay\\a\left(2-ay\right)-2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-ay\\2a-a^2y-2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2-ay\\y\left(a^2+2\right)=2a-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{2a-1}{a^2+2}\\x=2-ay=2-\dfrac{2a^2-a}{a^2+2}=\dfrac{2a^2+4-2a^2+a}{a^2+2}=\dfrac{a+4}{a^2+2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{a+4}{a^2+2}>0\\\dfrac{2a-1}{a^2+2}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a+4>0\\2a-1< 0\end{matrix}\right.\)
=>\(-4< a< \dfrac{1}{2}\)
giúp 2 bài vs làm b3 trc nhé
