ĐK:\(x\le2\)
\(\Leftrightarrow x^4-8x^2+16=x^2-4x+4\)
\(\Leftrightarrow\left(x^2-4\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)(TM)
ĐK:x\(\le2\)
\(\sqrt{x^4-8x^2+16}=2-x\Leftrightarrow\sqrt{\left(x^2\right)^2-2.x^2.4+4^2}=2-x\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x\Leftrightarrow\left|x^2-4\right|=2-x\Leftrightarrow\)(*)
Nếu \(-2\le x\le2\) thì (*)\(\Leftrightarrow4-x^2=2-x\Leftrightarrow\left(2-x\right)\left(2+x\right)-\left(2-x\right)=0\Leftrightarrow\left(2-x\right)\left(x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}2-x=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Nếu x<-2 thì (*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow\left(x-2\right)\left(x+2\right)=-\left(x-2\right)\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
Vậy S={-3;-1;2}
