a: Thay m=1 vào (I), ta được:
\(\left\{{}\begin{matrix}\left(1-2\right)x-3y=-5\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x-3y=-5\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x-3y+x+y=-5+3\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2y=-2\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=1\\x=3-1=2\end{matrix}\right.\)
b: Để hệ phương trình (I) có nghiệm duy nhất thì \(\dfrac{m-2}{1}\ne\dfrac{-3}{m}\)
=>\(m\left(m-2\right)\ne-3\)
=>\(m^2-2m+3\ne0\)
=>\(\left(m-1\right)^2+2\ne0\)(luôn đúng)
=>Hệ (I) luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}\left(m-2\right)x-3y=-5\\x+my=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m-2\right)x-3y=-5\\\left(m-2\right)x+\left(m^2-2m\right)y=\left(3m-6\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(m-2\right)x-3y-\left(m-2\right)x-\left(m^2-2m\right)y=-5-\left(3m-6\right)\\\left(m-2\right)x-3y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-3-m^2+2m\right)=-5-3m+6\\\left(m-2\right)x-3y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(-m^2+2m-3\right)=-3m+1\\\left(m-2\right)x-3y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m-1}{m^2-2m+3}\\3y=\left(m-2\right)x+5=\dfrac{\left(3m-1\right)\left(m-2\right)}{m^2-2m+3}+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m-1}{m^2-2m+3}\\3y=\dfrac{3m^2-7m+6+5m^2-10m+15}{m^2-2m+3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m-1}{m^2-2m+3}\\y=\dfrac{8m^2-17m+21}{3m^2-6m+9}\end{matrix}\right.\)
