a) \(\sqrt{9x^2}=3\)
⇔ \(\sqrt{\left(3x\right)^2}=3\)
\(\Rightarrow\) \(\left|3x\right|=3\)
\(\left[{}\begin{matrix}3x=3\\-3x=3\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a, \(\sqrt{9x^2}=3\)
\(\Leftrightarrow\) \(3x=3\)
\(\Leftrightarrow x=1\)
\(b,\sqrt{x^2-4x+4}+x=8\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)^2}+x=8\)
\(\Rightarrow\)\(x-2+x=8\)
\(\Leftrightarrow\)\(2x=10\)
\(\Leftrightarrow\)\(x=5\)
a) Ta có: \(\sqrt{9x^2}=3\)
\(\Leftrightarrow\left|3x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
b) Ta có: \(\sqrt{x^2-4x+4}+x=8\)
\(\Leftrightarrow\left|x-2\right|=8-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=8-x\left(x\ge2\right)\\x-2=x-8\left(x< 2\right)\end{matrix}\right.\Leftrightarrow x+x=8+2\)
\(\Leftrightarrow2x=10\)
hay x=5(nhận)
Vậy: x=5
