a: Thay a=3 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}3x-y=2\\x+3y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9x-3y=6\\x+3y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}10x=9\\3x-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{9}{10}\\y=3x-2=3\cdot\dfrac{9}{10}-2=\dfrac{27}{10}-2=\dfrac{7}{10}\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{a}{1}\ne\dfrac{-1}{a}\)
=>\(a^2\ne-1\)(luôn đúng)
vậy: Hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}ax-y=2\\x+ay=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=ax-2\\x+a\left(ax-2\right)=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(a^2+1\right)=2a+3\\y=ax-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2a+3}{a^2+1}\\y=\dfrac{2a^2+3a}{a^2+1}-2=\dfrac{2a^2+3a-2a^2-2}{a^2+1}=\dfrac{3a-2}{a^2+1}\end{matrix}\right.\)
