Question

[Lizy]

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

tìm số tự nhiên k để x+y=-1

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow\left\{{}\begin{matrix}k^2x-ky=2k\\x+ky=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2k+1}{k^2+1}\\y=\dfrac{2k^2+k}{k^2+1}-2=\dfrac{-k}{k^2+1}\end{matrix}\right.\)

\(x+y=-1\Rightarrow\dfrac{2k+1}{k^2+1}+\dfrac{-k}{k^2+1}=-1\)

\(\Rightarrow k+1=-k^2-1\)

\(\Rightarrow k^2+k+2=0\) (vô nghiệm)

Không tồn tại k thỏa mãn yêu cầu

Answer 2

[camcon]

Anh giúp em ạ! 

https://hoc24.vn/cau-hoi/limlimits-xrightarrow2-dfrac3x2x-12x2-5x2-cho-minh-hoi-khi-tu-duong-doi-voi-bai-nay-va-mau-dan-den-0-nhung-mau-lon-hon-0-hay-nho-hon-khong-theo-minh-hieu-la-gioi-han-dan-den-2.8768789368559