2) \(\sqrt{x^2-4x+4}=1\\ \Rightarrow x-2=1\\ \Rightarrow x=3\)
3) \(\sqrt{1-4x+4x^2}=5\\ \Rightarrow2x-1=5\\ \Rightarrow x=3\)
4) \(\sqrt{4\left(1-2x+x^2\right)}=0\\ \Rightarrow2\left(x-1\right)=0\\ \Rightarrow x-1=0\\ \Rightarrow x=1\)
5) \(\sqrt{9x^2}=2x+1\\ \Rightarrow2x+1-3x=0\\ \Rightarrow-x+=1\\ \Rightarrow x=1\)
10:ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2+8x+16}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
11: Ta có: \(\sqrt{9x^2+6x+1}=\sqrt{x^2-2\sqrt{6x}+6}\)
\(\Leftrightarrow\left|3x+1\right|=\left|x-\sqrt{6}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x-\sqrt{6}\\3x+1=-x+\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\sqrt{6}-1}{2}\\x=\dfrac{\sqrt{6}-1}{4}\end{matrix}\right.\)
