Question

[Nhân Tư]

\(x=\frac{2+k}{2k-1}\)

Thay vô pt2

\(k\cdot\frac{2+k}{2k-1}+y=k\)

\(\frac{2k+k^2}{2k-1}-k=-y\)

\(\frac{2k+k^2-2k^2+k}{2k-1}=-y\)

\(\frac{-k^2+3k}{2k-1}=-y\)

\(\frac{k^2-3k}{2k-1}=y\)

\(\Rightarrow x+y=\frac{2+k+k^2-3k}{2k-1}=\frac{k^2-2k+2}{2k-1}\)

Ta có: k²-2k+2=(k-1)²+1>0 mà x+y <0 nên 2k-1<0 =>k<1/2