Bài 2:
a: Ta có: \(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\left|2x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2+2x+2}=x-1\)
\(\Leftrightarrow x^2+2x+2=x^2-2x+1\)
\(\Leftrightarrow4x=-1\)
hay \(x=-\dfrac{1}{4}\left(loại\right)\)
Bài 3:
a: Ta có: \(A=\dfrac{12}{x-9}-\dfrac{\sqrt{x}+3}{x-3\sqrt{x}}\)
\(=\dfrac{12\sqrt{x}-x-6\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-x+6\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
