Để hai đường thẳng này song song thì
\(\left\{{}\begin{matrix}2-k^2=k\\k-5< >3k-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-k^2-k+2=0\\-2k\ne-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}k^2+k-2=0\\k\ne1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(k+2\right)\left(k-1\right)=0\\k\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k\in\left\{-2;1\right\}\\k\ne1\end{matrix}\right.\)
=>k=-2
ĐKXĐ: \(\left\{{}\begin{matrix}2-k^2\ne0\\k\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}k\ne0\\k\ne\sqrt{2}\\k\ne-\sqrt{2}\end{matrix}\right.\)
Để hai đường thẳng đã cho song song thì:
\(\left\{{}\begin{matrix}2-k^2=k\\k-5\ne3k-7\end{matrix}\right.\)
*) \(2-k^2=k\)
\(\Leftrightarrow k^2+k-2=0\)
\(\Leftrightarrow k^2-k+2k-2=0\)
\(\Leftrightarrow\left(k^2-k\right)+\left(2k-2\right)=0\)
\(\Leftrightarrow k\left(k-1\right)+2\left(k-1\right)=0\)
\(\Leftrightarrow\left(k-1\right)\left(k+2\right)=0\)
\(\Leftrightarrow k-1=0;k+2=0\)
+) \(k-1=0\)
\(\Leftrightarrow k=1\) (nhận) (1)
+) \(k+2=0\)
\(\Leftrightarrow k=-2\) (nhận) (2)
*) \(k-5\ne3k-7\)
\(\Leftrightarrow k-3k\ne-7+5\)
\(\Leftrightarrow-2k\ne-2\)
\(\Leftrightarrow k\ne1\) (3)
Từ (1), (2) và (3) \(\Rightarrow k=-2\) thì hai đường thẳng đã cho song song
