Question

\(\left\{{}\begin{matrix}kx-y=2\\x+ky=1\end{matrix}\right.\)

a) giải hpt khi k=5

b)gọi nghiệm của hpt là (x;y) . Tìm số tự nhiên k để x+y=-1

Answer 1

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 ....

Answer 2

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 .