Question

\(\sqrt{4x^2-12x+9}=x\)

Answer 1

\(ĐK:x\ge0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x\left(x\ge\dfrac{3}{2}\right)\\2x-3=-x\left(0\le x< \dfrac{3}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

Answer 2

2x - 3 - x = 0

x = 3

Answer 3

\(\sqrt{4x^2-12x+9}=x\)

\(< =>\left|2x-3\right|=x\\ < =>TH1:2x-3\le0=>x\le\dfrac{3}{2}\\ ->3-2x=x\\ < =>x=1\left(tmdk\right)\\ TH2:2x-3>0< =>x>\dfrac{3}{2}\\ ->2x-3=x\\ < =>x=3\left(tmdk\right)\)

=> Ptrinhf có tập nghiệm: S={1;3}

Answer 4

ĐK: \(x\ge0\)

Ta có: \(\sqrt{4x^2-12x+9}=x\)

      \(\Leftrightarrow4x^2-12x+9=x^2\)

      \(\Leftrightarrow3x^2-12x+9=0\)

      \(\Leftrightarrow x^2-4x+3=0\)

      \(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)