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Lời giải:
a)
Khi $m=1$ thì HPT trở thành:\(\left\{\begin{matrix} x-y=2\\ x+y=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2x=2+1\\ 2y=1-2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{3}{2}\\ y=\frac{-1}{2}\end{matrix}\right.\)
b)
HPT \(\Leftrightarrow \left\{\begin{matrix} mx-y=2\\ x=1-my\end{matrix}\right.\Rightarrow m(1-my)-y=2\)
\(\Leftrightarrow y(m^2+1)=m-2\Rightarrow y=\frac{m-2}{m^2+1}\)
\(x=1-my=1-\frac{m^2-2m}{m^2+1}=\frac{1+2m}{m^2+1}\)
Để $x+y=-1$
$\Leftrightarrow \frac{m-2}{m^2+1}+\frac{1+2m}{m^2+1}=-1$
$\Leftrightarrow \frac{3m-1}{m^2+1}=-1$
$\Rightarrow 3m-1=-m^2-1$
$\Leftrightarrow m^2+3m=0\Rightarrow m=0$ hoặc $m=-3$
\(\Leftrightarrow\left\{{}\begin{matrix}k^2x-ky=2k\\x+ky=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(k^2+1\right)x=2k+1\\y=kx-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2k+1}{k^2+1}\\y=\dfrac{2k^2+k}{k^2+1}-2=\dfrac{-k}{k^2+1}\end{matrix}\right.\)
\(x+y=-1\Rightarrow\dfrac{2k+1}{k^2+1}+\dfrac{-k}{k^2+1}=-1\)
\(\Rightarrow k+1=-k^2-1\)
\(\Rightarrow k^2+k+2=0\) (vô nghiệm)
Không tồn tại k thỏa mãn yêu cầu
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:
\(\left\{{}\begin{matrix}x=\frac{\left(2m+1\right)\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\frac{2m+1}{m+1}\\y=2m-mx=\frac{m}{m+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x+ky=1\\kx+2y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\k\left(-\dfrac{k}{2}y+\dfrac{1}{2}\right)+2y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\end{matrix}\right.\)
Hệ PT có nghiệm \(\Leftrightarrow\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\) có nghiệm
\(\Leftrightarrow-\dfrac{k^2}{2}+2\ne0\Leftrightarrow\dfrac{k^2}{2}=2\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Bài 1.
\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=m-1\end{matrix}\right.\)
\(x_0^2+y_0^2=9m\)
\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)
\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)
\(\Leftrightarrow2m^2-7m+5=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
Do \(x=2\) là nghiệm của phương trình nên:
\(\left\{{}\begin{matrix}2a+y=3\\2+ay=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=3-2a\\ay=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ay=3a-2a^2\\ay=-3\end{matrix}\right.\)
\(\Rightarrow3a-2a^2=-3\)
\(\Rightarrow2a^2-3a-3=0\Rightarrow a=\dfrac{3\pm\sqrt{33}}{4}\)
a) \(\left\{{}\begin{matrix}5x-y=2\\x+5y=1\end{matrix}\right.\)
-> \(\left\{{}\begin{matrix}5x-y=2\\5x+25y=5\end{matrix}\right.\)
->\(\left\{{}\begin{matrix}26y=3\\5x-y=2\end{matrix}\right.\)
->\(\left\{{}\begin{matrix}y=\frac{3}{26}\\x=\frac{11}{26}\end{matrix}\right.\)
vậy...
b)\(\left\{{}\begin{matrix}x+y=-1\\kx-y=2\\x+ky=1\end{matrix}\right.\)
->\(\left\{{}\begin{matrix}y=-1-x\\kx-y=2\\x+ky=1\end{matrix}\right.\)
->\(\left\{{}\begin{matrix}kx-\left(-1-x\right)=2\\x+k\left(-1-x\right)=1\end{matrix}\right.\)
->\(\left\{{}\begin{matrix}x\left(k+1\right)=1\\x\left(1-k\right)=1+k\end{matrix}\right.\)
->\(\left\{{}\begin{matrix}x=\frac{1}{k+1}\\x=\frac{1+k}{1-k}\end{matrix}\right.\) dk x\(\ne\)-1 ; x\(\ne\)1
->\(\frac{1}{k+1}=\frac{1+k}{1-k}\)
->\(1-k=k^2+2k+1\)
->k2+3k=0
->\(\left[{}\begin{matrix}k=-3\\k=0\end{matrix}\right.\)(nhận)
vậy ....
a, Thay k = 5 vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}5x-y=2\\x+5y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}25x-5y=10\\x+5y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x-y=2\\26x=11\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{55}{26}-y=2\\x=\frac{11}{26}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=\frac{3}{26}\\x=\frac{11}{26}\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm \(\left(x;y\right)=\left(\frac{11}{26};\frac{3}{26}\right)\) với giá trị của k = 5 .
b, Ta có : \(\left\{{}\begin{matrix}kx-y=2\\x+ky=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=kx-2\\x+k\left(kx-2\right)=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=kx-2\\x+k^2x-2k=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=kx-2\\x\left(k^2+1\right)=1+2k\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=\frac{k\left(1+2k\right)}{k^2+1}-2\\x=\frac{1+2k}{k^2+1}\end{matrix}\right.\)
- Để \(x+y=-1\) thì :
\(\frac{1+2k}{k^2+1}+\frac{k\left(1+2k\right)}{k^2+1}-2=-1\)
=> \(\frac{k\left(1+2k\right)+1+2k}{k^2+1}=1\)
=> \(k\left(1+2k\right)+1+2k=k^2+1\)
=> \(k+2k^2+1+2k-k^2-1=0\)
=> \(k^2+3k=0\)
=> \(\left[{}\begin{matrix}k=0\\k=-3\end{matrix}\right.\)
Vậy để thỏa mãn điều kiền trên thì k có giá trị là 0 hay -3 .