Question

Giải phương trình

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

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-4x-5\right)^2=\left(4x-17\right)^2\\\left(x-5\right)\left(x+1\right)>=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-4x-5-4x+17\right)\left(x^2-4x-5+4x-17\right)=0\\x\in(-\infty;-1]\cup[5;+\infty)\end{matrix}\right.\)

 

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-8x+12\right)\left(x^2-22\right)=0\\x\in(-\infty;-1]\cup[5;+\infty)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x-6\right)\left(x^2-22\right)=0\\x\in(-\infty;-1]\cup[5;+\infty)\end{matrix}\right.\Leftrightarrow x\in\left\{6;-\sqrt{22}\right\}\)