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\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=m-y\\m-y+ym+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m-y\\ym=1-m\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=m-\dfrac{1-m}{m}=\dfrac{m^2+m-1}{m}\\y=\dfrac{1-m}{m}\end{matrix}\right.\)
\(x+2y>0\\ \Leftrightarrow\dfrac{m^2+m-1}{m}+\dfrac{2-2m}{m}>0\\ \Leftrightarrow\dfrac{m^2-m+1}{m}>0\)
Mà \(m^2-m+1=\left(m-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
Vậy \(m>0\) thỏa đề
a. Thay k=5, ta có hpt:
\(\left\{{}\begin{matrix}5x-y=2\\x+5y=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{11}{26}\\y=\dfrac{3}{26}\end{matrix}\right.\)
Vậy hpt có nghiệm là \(\left(\dfrac{11}{26};\dfrac{3}{26}\right)\)
b.ĐK: \(k\ne-\dfrac{1}{k}\)\(\Leftrightarrow k\forall R\)
hpt\(\Leftrightarrow\left\{{}\begin{matrix}kx-y=2\left(1\right)\\kx+k^2y=k\left(2\right)\end{matrix}\right.\)
Trừ hai pt, ta được: \(\left(k^2+1\right)y=k-2\)\(\Leftrightarrow y=\dfrac{k-2}{k^2+1}\)
Thay vào (1), ta có: \(kx=2+\dfrac{k-2}{k^2+1}\)\(\Leftrightarrow x=\dfrac{2k^2+k}{k^3+k}\)\(=\dfrac{2k+1}{k^2+1}\)
\(x+y=\dfrac{3k-1}{k^2+1}\)
\(\dfrac{3k-1}{k^2+1}=\dfrac{-3}{k^2+1}\)
\(\Rightarrow k=\dfrac{-2}{3}\)
\(\left\{{}\begin{matrix}2x+ky=1\\kx+2y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\k\left(-\dfrac{k}{2}y+\dfrac{1}{2}\right)+2y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\end{matrix}\right.\)
Hệ PT có nghiệm \(\Leftrightarrow\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\) có nghiệm
\(\Leftrightarrow-\dfrac{k^2}{2}+2\ne0\Leftrightarrow\dfrac{k^2}{2}=2\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Để hệ có nghiệm duy nhất thì \(\dfrac{2}{m}\ne\dfrac{-1}{1}=-1\)
=>\(m\ne-2\)
\(\left\{{}\begin{matrix}2x-y=1\\mx+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y+mx+y=6\\2x-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m+2\right)=6\\y=2x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6}{m+2}\\y=2\cdot\dfrac{6}{m+2}-1=\dfrac{12}{m+2}-1=\dfrac{12-m-2}{m+2}=\dfrac{-m+10}{m+2}\end{matrix}\right.\)
Để x>0 và y<0 thì \(\left\{{}\begin{matrix}\dfrac{6}{m+2}>0\\\dfrac{-m+10}{m+2}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m+2>0\\\dfrac{m-10}{m+2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-2\\\left[{}\begin{matrix}m>10\\m< -2\end{matrix}\right.\end{matrix}\right.\)
=>m>10
Lời giải:
$x+2y=5\Leftrightarrow x=5-2y$. Thay vô pt $(1)$
$m(5-2y)+y=4$
$\Leftrightarrow y(1-2m)=4-5m$
Để pt có nghiệm duy nhất thì $1-2m\neq 0\Leftrightarrow m\neq \frac{1}{2}$
Khi đó: $y=\frac{4-5m}{1-2m}$
$x=5-2y=5-\frac{2(4-5m)}{1-2m}=\frac{-3}{1-2m}$
$x>0\Leftrightarrow \frac{-3}{1-2m}>0\Leftrightarrow 1-2m<0\Leftrightarrow m> \frac{1}{2}(1)$
$y>0\Leftrightarrow \frac{4-5m}{1-2m}>0\Leftrightarrow 4-5m<0$ (do $1-2m< 0$
$\Leftrightarrow m> \frac{4}{5}(2)$
Từ $(1); (2)\Rightarrow m> \frac{4}{5}$
$x> y\Leftrightarrow \frac{-3}{1-2m}> \frac{4-5m}{1-2m}$
$\Leftrightarrow \frac{5m-7}{1-2m}>0$
$\Leftrightarrow 5m-7< 0$ (do $1-2m<0$)
$\Leftrightarrow m< \frac{7}{5}$
Vậy $\frac{4}{5}< m< \frac{7}{5}$
Lời giải:
$x+2y=5\Leftrightarrow x=5-2y$. Thay vô pt $(1)$
$m(5-2y)+y=4$
$\Leftrightarrow y(1-2m)=4-5m$
Để pt có nghiệm duy nhất thì $1-2m\neq 0\Leftrightarrow m\neq \frac{1}{2}$
Khi đó: $y=\frac{4-5m}{1-2m}$
$x=5-2y=5-\frac{2(4-5m)}{1-2m}=\frac{-3}{1-2m}$
$x>0\Leftrightarrow \frac{-3}{1-2m}>0\Leftrightarrow 1-2m<0\Leftrightarrow m> \frac{1}{2}(1)$
$y>0\Leftrightarrow \frac{4-5m}{1-2m}>0\Leftrightarrow 4-5m<0$ (do $1-2m< 0$
$\Leftrightarrow m> \frac{4}{5}(2)$
Từ $(1); (2)\Rightarrow m> \frac{4}{5}$
$x> y\Leftrightarrow \frac{-3}{1-2m}> \frac{4-5m}{1-2m}$
$\Leftrightarrow \frac{5m-7}{1-2m}>0$
$\Leftrightarrow 5m-7< 0$ (do $1-2m<0$)
$\Leftrightarrow m< \frac{7}{5}$
Vậy $\frac{4}{5}< m< \frac{7}{5}$
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{1}\)
=>\(m^2\ne1\)
=>\(m\notin\left\{1;-1\right\}\)
Khi \(m\notin\left\{1;-1\right\}\) thì \(\left\{{}\begin{matrix}x+my=m+1\\mx+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m\left(m+1-my\right)+y=2m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=m+1-my\\m^2+m-m^2y+y-2m=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-m^2+1\right)=-m^2+m\\x=m+1-my\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m^2-m}{m^2-1}=\dfrac{m\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\dfrac{m}{m+1}\\x=m+1-\dfrac{m^2}{m+1}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m}{m+1}\\x=\dfrac{\left(m+1\right)^2-m^2}{m+1}=\dfrac{2m+1}{m+1}\end{matrix}\right.\)
Để \(\left\{{}\begin{matrix}x>=2\\y>=1\end{matrix}\right.\) thì \(\left\{{}\begin{matrix}\dfrac{2m+1}{m+1}>=2\\\dfrac{m}{m+1}>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2\left(m+1\right)}{m+1}>=0\\\dfrac{m-m-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2m+1-2m-2}{m+1}>=0\\\dfrac{-1}{m+1}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{m+1}>=0\\-\dfrac{1}{m+1}>=0\end{matrix}\right.\Leftrightarrow m+1< 0\)
=>m<-1