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a) Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{4}\ne\dfrac{-1}{-m}\)
\(\Leftrightarrow-m^2\ne-4\)
\(\Leftrightarrow m^2\ne4\)
hay \(m\notin\left\{2;-2\right\}\)
1:
a)\(\hept{\begin{cases}nx+x=5
\\x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x.\left(n+1\right)=5\left(1\right)\\x+y=1\end{cases}}\)
Để hệ vô nghiệm thì \(\dfrac{m}{4}=\dfrac{-1}{-m}< >\dfrac{2m}{m+6}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{4}=\dfrac{1}{m}\\\dfrac{1}{m}< >\dfrac{2m}{m+6}\\\dfrac{m}{4}< >\dfrac{2m}{m+6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=4\\2m^2< >m+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\2m^2-m-6< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left(m-2\right)\left(2m+3\right)< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m\notin\left\{2;-\dfrac{3}{2}\right\}\end{matrix}\right.\Leftrightarrow m=-2\)
Để hệ vô số nghiệm thì \(\dfrac{m}{4}=\dfrac{-1}{-m}=\dfrac{2m}{m+6}\)
=>\(\left\{{}\begin{matrix}\dfrac{m}{4}=\dfrac{1}{m}\\\dfrac{1}{m}=\dfrac{2m}{m+6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m^2=4\\2m^2=m+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\2m^2-m-6=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\2m^2-4m+3m-6=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left(m-2\right)\left(2m+3\right)=0\end{matrix}\right.\Leftrightarrow m=2\)
\(\left\{{}\begin{matrix}m^2x+my=m\\x+my=m+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-1\right)x=-1\\x+my=m+1\end{matrix}\right.\)
- Với \(m=\pm1\Rightarrow0.x=-1\) hệ vô nghiệm
- Không tồn tại m để hệ có vô số nghiệm
- Với \(m\ne\pm1\) hệ có nghiệm duy nhất
Câu 3:
\(\left\{{}\begin{matrix}mx+4y=9\\mx+m^2y=8m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}mx+4y=9\\\left(m^2-4\right)y=8m-9\end{matrix}\right.\)
Để hpt đã cho có nghiệm \(\Leftrightarrow m\ne\pm2\)
Khi đó ta có: \(\left\{{}\begin{matrix}y=\frac{8m-9}{m^2-4}\\x=8-my=8-\frac{8m^2-9m}{m^2-4}=\frac{9m-32}{m^2-4}\end{matrix}\right.\)
\(2x+y+\frac{38}{m^2-4}=3\)
\(\Leftrightarrow\frac{18m-64}{m^2-4}+\frac{8m-9}{m^2-4}+\frac{38}{m^2-4}=3\)
\(\Leftrightarrow26m-35=3m^2-12\)
\(\Leftrightarrow3m^2-26m+23=0\Rightarrow\left[{}\begin{matrix}m=1\\m=\frac{23}{3}\end{matrix}\right.\)
Câu 4:
\(\left\{{}\begin{matrix}m^2x-my=2m^2\\4x-my=m+6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-4\right)x=2m^2-m-6\\4x-my=m+6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)\left(m+2\right)x=\left(m-2\right)\left(2m+3\right)\\4x-my=m+6\end{matrix}\right.\)
- Với \(m=-2\) hệ vô nghiệm
- Với \(m=2\) hệ có vô số nghiệm thỏa mãn \(2x-y=4\)
- Với \(m\ne\pm2\) hệ có nghiệm duy nhất:
\(\left\{{}\begin{matrix}x=\frac{2m+3}{m+2}\\y=mx-2m=\frac{2m^2+3m-2m^2-4m}{m+2}=\frac{-m}{m+2}\end{matrix}\right.\)
Câu 1: ĐKXĐ \(\left\{{}\begin{matrix}x\ne1\\y\ne-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=u\\\frac{1}{y+1}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2u+v=7\\5u-2v=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4u+2v=14\\5u-2v=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u=2\\v=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=2\\\frac{1}{y+1}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=\frac{1}{2}\\y+1=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{3}{2}\\y=-\frac{2}{3}\end{matrix}\right.\)
Câu 2:
Để hệ có nghiệm (x;y)=\(\left(2;-1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2m.2-\left(m+1\right).\left(-1\right)=m-n\\\left(m+2\right).2+3n\left(-1\right)=2m-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m+n=-1\\3n=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n=\frac{7}{3}\\m=\frac{5}{6}\end{matrix}\right.\)
c) Để hệ phương trình vô nghiệm thì \(\dfrac{m}{4}=\dfrac{-1}{-m}\ne\dfrac{2m}{6+m}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m}{4}=\dfrac{1}{m}\\\dfrac{m}{4}\ne\dfrac{2m}{6+m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=4\\m\left(m+6\right)\ne8m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m^2+6m-8m\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m^2-2m\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m\left(m-2\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left\{{}\begin{matrix}m\ne0\\m-2\ne0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left\{{}\begin{matrix}m\ne0\\m\ne2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m=-2\)
b) Để hệ phương trình có vô số nghiệm thì \(\dfrac{m}{4}=\dfrac{-1}{-m}=\dfrac{2m}{6+m}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m}{4}=\dfrac{1}{m}\\\dfrac{m}{4}=\dfrac{2m}{6+m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=4\\m\left(6+m\right)=8m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\6m+m^2-8m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m^2-2m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\m\left(m-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left[{}\begin{matrix}m=0\\m-2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{2;-2\right\}\\\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m=2\)