Question

Giải phương trình

\(\left|x^2-4x-5\right|=4x-17\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{17}{4}\\\left(x^2-4x-5\right)^2=\left(4x-17\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{17}{4}\\\left(x^2-4x-5-4x+17\right)\left(x^2-4x-5+4x-17\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{17}{4}\\\left(x^2-8x+12\right)\left(x^2-22\right)=0\end{matrix}\right.\Leftrightarrow x=6\)