Question

giải phương trình 

\(\sqrt{x^2-4x+4}-2x+5=0\)

Answer 1

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=2x-5\)

\(\Leftrightarrow\left|x-2\right|=2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2x-5\left(x\ge2\right)\\x-2=5-2x\left(\dfrac{5}{2}\le x< 2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\dfrac{7}{3}\left(ktm\right)\end{matrix}\right.\)

Answer 2

\(ĐK:x\ge\dfrac{5}{2}\\ PT\Leftrightarrow\left|x-2\right|=2x-5\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-5\left(x\ge2\right)\\x-2=5-2x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\dfrac{7}{3}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=3\)