1: Ta có: \(\left|x^2-72\right|-2=70\)
\(\Leftrightarrow\left|x^2-72\right|=72\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-72=72\\x^2-72=-72\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=144\\x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\\x=0\end{matrix}\right.\)
Vậy: S={0;-12;12}
2: Ta có: \(x^2-5\left|x\right|=14\)
\(\Leftrightarrow5\left|x\right|=x^2-14\)(*)
Trường hợp 1: x≥0
(*)\(\Leftrightarrow5x=x^2-14\)
\(\Leftrightarrow x^2-14=5x\)
\(\Leftrightarrow x^2-5x-14=0\)
\(\Leftrightarrow x^2-7x+2x-14=0\)
\(\Leftrightarrow x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Trường hợp 2: x<0
(*)\(\Leftrightarrow-5x=x^2-14\)
\(\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow x^2+7x-2x-14=0\)
\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
Vậy: S={-7;7}
