Question

\(\sqrt{\left(x^2-4x+4\right)}=2x-3\)

Answer 1

ĐK : \(x\ge\dfrac{3}{2}\)

Lời giải ..............

\(\sqrt{x^2-4x+4}=2x-3\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}-\left(2x-3\right)=0\)

\(\Leftrightarrow\left|x-2\right|-2x+3=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-2\right|-2x+3=0\\\left|2-x\right|-2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+1=0\\5-3x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\left(loại\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{5}{3}\right\}\)

Answer 2

Đk : \(\forall x\in R\)

Pt

<=> \(\sqrt{\left(x-2\right)^2}=2x-3\)

<=>\(x-2=2x-3\)

\(\Leftrightarrow-x=-1\)

<=> x = 1

Vậy S={1}