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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 .
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 $m=-3$
\(\left\{{}\begin{matrix}2x+ky=1\\kx+2y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\k\left(-\dfrac{k}{2}y+\dfrac{1}{2}\right)+2y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{k}{2}y+\dfrac{1}{2}\\\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\end{matrix}\right.\)
Hệ PT có nghiệm \(\Leftrightarrow\left(-\dfrac{k^2}{2}+2\right)y+\left(\dfrac{k}{2}-1\right)=0\) có nghiệm
\(\Leftrightarrow-\dfrac{k^2}{2}+2\ne0\Leftrightarrow\dfrac{k^2}{2}=2\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)