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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
\(\left\{{}\begin{matrix}5x=5m\\y=2x-m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=10-m+1=11-m\end{matrix}\right.\)
Thay vào ta đc
\(2m^2-3\left(11-m\right)=2\Leftrightarrow2m^2-33+3m=2\Leftrightarrow2m^2+3m-35=0\Leftrightarrow m=\dfrac{7}{2};m=-5\)
1. \(\Leftrightarrow\left\{{}\begin{matrix}mx+m^2y=3m\\mx+4y=6\end{matrix}\right.\)
\(\Rightarrow\left(m^2-4\right)y=3\left(m-2\right)\)
\(\Leftrightarrow\left(m-2\right)\left(m+2\right)y=3\left(m-2\right)\)
Để pt có nghiệm duy nhất \(\Rightarrow\left(m-2\right)\left(m+2\right)\ne0\Rightarrow m\ne\pm2\)
Để pt vô nghiệm \(\Rightarrow\left\{{}\begin{matrix}\left(m-2\right)\left(m+2\right)=0\\3\left(m-2\right)\ne0\end{matrix}\right.\) \(\Rightarrow m=-2\)
2. Không thấy m nào ở hệ?
3. Bạn tự giải câu a
b/ \(\left\{{}\begin{matrix}6x+2my=2m\\\left(m^2-m\right)x+2my=m^2-m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\frac{\left(m-1\right)\left(1-x\right)}{2}\\\left(m^2-m-6\right)x=m^2-3m\end{matrix}\right.\)
Để hệ có nghiệm duy nhất \(\Rightarrow m^2-m-6\ne0\Rightarrow m\ne\left\{-2;3\right\}\)
Khi đó: \(\left\{{}\begin{matrix}x=\frac{m^2-3m}{m^2-m-6}=\frac{m}{m+2}\\y=\frac{\left(m-1\right)\left(1-x\right)}{2}=\frac{m-1}{m+2}\end{matrix}\right.\)
\(x+y^2=1\Leftrightarrow\frac{m}{m+2}+\frac{\left(m-1\right)^2}{\left(m+2\right)^2}=1\)
\(\Leftrightarrow m\left(m+2\right)+\left(m-1\right)^2=\left(m+2\right)^2\)
\(\Leftrightarrow m^2-4m-3=0\Rightarrow\) bấm máy, số xấu
4.
\(\Leftrightarrow\left\{{}\begin{matrix}m^2x+my=2m^2\\x+my=m+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-1\right)x=2m^2-m-1=\left(2m+1\right)\left(m-1\right)\\y=2m-mx\end{matrix}\right.\)
- Với \(m=1\) hệ có vô số nghiệm
- Với \(m=-1\) hệ vô nghiệm
- Với \(m\ne\pm1\) hệ có nghiệm duy nhất:
\(\left\{{}\begin{matrix}x=\frac{\left(2m+1\right)\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\frac{2m+1}{m+1}\\y=2m-mx=\frac{m}{m+1}\end{matrix}\right.\)
a)
Khi m = 1, ta có:
{ x+2y=1+3
2x-3y=1
=> { x+2y=4
2x-3y=1
=> { 2x+4y=8
2x-3y=1
=> { x+2y=4
2x-3y-2x-4y=1-8
=> { x=4-2y
-7y = -7
=> { x = 2
y = 1
Vậy khi m = 1 thì hệ phương trình có cặp nghệm
(x; y) = (2;1)
a) Thay m=1 vào HPT ta có:
\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}2x+4y=8\\7y=7\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y)= (2;1)
a) Thay m=1 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7y=7\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=4-2y=4-2=2\end{matrix}\right.\)
Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất là (x,y)=(2;1)
b) Ta có: \(\left\{{}\begin{matrix}x+2y=m+3\\2x-3y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\2\left(m+3-2y\right)-3y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\2m+6-4y-3y-m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\-7y+m+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\-7y=-m-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2y\\y=\dfrac{m+6}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+3-2\cdot\dfrac{m+6}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+3-\dfrac{2m+12}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7m+21-2m-12}{7}=\dfrac{5m+9}{7}\\y=\dfrac{m+6}{7}\end{matrix}\right.\)
Để hệ phương trình có nghiệm duy nhất thỏa mãn x+y=3 thì \(\dfrac{5m+9}{7}+\dfrac{m+6}{7}=3\)
\(\Leftrightarrow6m+15=21\)
\(\Leftrightarrow6m=6\)
hay m=1
Vậy: Khi m=1 thì hệ phương trình có nghiệm duy nhất thỏa mãn x+y=3
a/ Thay \(m=1\) vào hpt ta có :
\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy...
b/ Ta có :
\(\left\{{}\begin{matrix}x+2y=m+3\\2x-3y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{2y}\\\dfrac{2\left(m+3\right)}{2y}-3y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{2y}\\\dfrac{m+3}{y}-3y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{2y}\\m-3y^2+3=my\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+my=3\\mx-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-my\\m\left(3-my\right)-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3m-m^2y-3y=1\\x=3-my\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\left(m^2+3\right)=3m-1\\x=3-my\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3m-1}{m^2+3}\\x=3-\frac{m\left(3m-1\right)}{m^2+3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{m+9}{m^2+3}\\y=\frac{3m-1}{m^2+3}\end{matrix}\right.\)
Khi đó: \(x+y=\frac{m+9+3m-1}{m^2+3}=1\)
\(\Leftrightarrow4m+8=m^2+3\)
\(\Leftrightarrow m^2-4m-5=0\)
\(\Leftrightarrow\left(m-5\right)\left(m+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=5\\m=-1\end{matrix}\right.\)( thỏa mãn )
Vậy....
@Nguyễn Ngọc Lộc