1
ĐK: \(x\in R\)
\(\sqrt{x^2-4x+4}=\sqrt{4x^2-12+9}\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\\ \Leftrightarrow\left|x-2\right|=\left|2x-3\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\\2-x=2x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
2
ĐK: \(\left\{{}\begin{matrix}x+2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)
Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow t^2=x-1\Rightarrow x=t^2+1\)
\(\sqrt{x+2\sqrt{x-1}}=2\\ \Leftrightarrow\sqrt{t^2+2t+1}=2\\ \Leftrightarrow\sqrt{\left(t+1\right)^2}=2\left(1\right)\)
Do có \(t\ge0\) nên \(\left(1\right)\Leftrightarrow t+1=2\Leftrightarrow t=2-1=1\)
\(\Rightarrow x=t^2+1=1^2+1=2\) (thỏa mãn)
1: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc 3x=5
=>x=5/3 hoặc x=1
2: \(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
=>căn x-1+1=2
=>căn x-1=1
=>x-1=1
=>x=2
