a: Thay m=0 vào (I), ta được:
\(\left\{{}\begin{matrix}2x+0\cdot y=5\\3x-y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=5\\y=3x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=3\cdot\dfrac{5}{2}=\dfrac{15}{2}\end{matrix}\right.\)
b: Để hệ (I) có nghiệm duy nhất thì \(\dfrac{2}{3}\ne\dfrac{m}{-1}\)
=>\(m\ne-\dfrac{2}{3}\)
\(\left\{{}\begin{matrix}2x+my=5\\3x-y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3x\\2x+m\cdot3x=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3x\\x\left(3m+2\right)=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{3m+2}\\y=3\cdot\dfrac{5}{3m+2}=\dfrac{15}{3m+2}\end{matrix}\right.\)
\(x-y+\dfrac{m+1}{m-2}=-4\)
=>\(\dfrac{10}{3m+2}+\dfrac{m+1}{m-2}=-4\)
=>\(\dfrac{10\left(m-2\right)+\left(m+1\right)\left(3m+2\right)}{\left(3m+2\right)\left(m-2\right)}=-4\)
=>\(10m-20+3m^2+2m+3m+2=-4\left(3m^2-6m+2m-4\right)\)
=>\(3m^2+15m-18=-4\left(3m^2-4m-4\right)\)
=>\(3m^2+15m-18+12m^2-16m-16=0\)
=>\(15m^2-m-34=0\)(1)
\(\text{Δ}=\left(-1\right)^2-4\cdot15\cdot\left(-34\right)=2041>0\)
=>Phương trình (1) có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}m_1=\dfrac{1-\sqrt{2041}}{30}\left(nhận\right)\\m_2=\dfrac{1+\sqrt{2041}}{30}\left(nhận\right)\end{matrix}\right.\)
