18:
Để hệ có nghiệm duy nhất thì \(\dfrac{2}{m-1}\ne\dfrac{3}{2m}\)
=>\(4m\ne3m-3\)
=>\(m\ne-3\)
\(\left\{{}\begin{matrix}2x+3y=8\\\left(m-1\right)x+2my=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left(m-1\right)x+\left(3m-3\right)y=8m-8\\2\left(m-1\right)x+4my=14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2m-2\right)x+\left(3m-3\right)y-\left(2m-2\right)x-4my=8m-8-14\\2x+3y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(-m-3\right)y=8m-22\\2x=8-3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-8m+22}{m+3}\\x=4-\dfrac{3}{2}y=4-\dfrac{3}{2}\cdot\dfrac{-8m+22}{m+3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-8m+22}{m+3}\\x=4-3\cdot\dfrac{-4m+11}{m+3}=4-\dfrac{-12m+33}{m+3}=\dfrac{4m+12+12m-33}{m+3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{-8m+22}{m+3}\\x=\dfrac{16m-21}{m+3}\end{matrix}\right.\)
3x-y=1
=>\(\dfrac{3\left(16m-21\right)-\left(-8m+22\right)}{m+3}=1\)
=>48m-63+8m-22=m+3
=>56m-85=m+3
=>55m=88
=>m=1,6(nhận)
20: Vì \(\dfrac{2}{-1}\ne\dfrac{-5}{3}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}2x-5y=m-1\\-x+3y=-m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-5y=m-1\\-2x+6y=-2m+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5y-2x+6y=m-1-2m+4\\-x+3y=-m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-m+3\\-x=-m+2-3y=-m+2-3\left(-m+3\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-m+3\\-x=-m+2+3m-9=2m-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2m+7\\y=-m+3\end{matrix}\right.\)
\(E=x^2+y^2=\left(-2m+7\right)^2+\left(-m+3\right)^2\)
\(=4m^2-28m+49+m^2-6m+9\)
\(=5m^2-34m+58\)
\(=5\left(m^2-\dfrac{34}{5}m+\dfrac{58}{5}\right)\)
\(=5\left(m^2-2\cdot m\cdot\dfrac{17}{5}+\dfrac{289}{25}+\dfrac{1}{25}\right)\)
\(=5\left(m-\dfrac{17}{5}\right)^2+\dfrac{1}{5}>=\dfrac{1}{5}\forall m\)
Dấu '=' xảy ra khi \(m-\dfrac{17}{5}=0\)
=>\(m=\dfrac{17}{5}\)