Ta có :
\(\sqrt{4\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow\sqrt{4\left(1-x\right)^2}=6\)
\(\Leftrightarrow4\left(1-x\right)^2=36\)
\(\Leftrightarrow\left(1-x\right)^2=9\)
\(\Leftrightarrow1-2x+x^2=9\)
\(\Leftrightarrow x^2-2x+1-9=0\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
Vậy....
Điều kiện: \(x\in R\)
Ta có:\(\sqrt{4\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow\sqrt{\left(2-2x\right)^2}=6\)
\(\Leftrightarrow\left|2-2x\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2-2x=6\\2-2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=4\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-2;4}
ĐKXĐ : \(x\le1\)
Pt đã cho tương đương :
\(\sqrt{4\left(1-x\right)^2}=6\)
\(\Leftrightarrow4.\left(1-x\right)^2=36\)
\(\Leftrightarrow\left(x-1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\) Mà \(x\le1\)
$\to x=-2$ thỏa mãn đề.
