Đáp án:
a)Thay k=5 hệ phương trình kx-y=2 và x+ky=1 ta có:
$\left \{ {{5x-y=2} \atop {x+5y=1}} \right.$
<=> $\left \{ {{5x-y=2} \atop {5x+25y=5}} \right.$
<=> $\left \{ {{5x-y=2} \atop {-26y=-3}} \right.$
=>y=3/26;
=>x=11/26;
b) $\left \{ {{kx-y=2} \atop {x+ky=1}} \right.$ ;
$\left \{ {{kx-y=2} \atop {kx+k^{2}y=k}} \right.$
$\left \{ {{kx-y=2} \atop {(-1-k^{2})y=k}} \right.$
=> y=$\frac{k}{-1-k^{2}}$
=>x= $\frac{2-k+2k^{2}}{k+k^{3}}$
x+y=$\frac{2-k+2k^{2}}{k+k^{3}}$ + $\frac{k}{-1-k^{2}}$=-1