a) \(3x-1-\sqrt{4x^2-12x+9}=0\)
\(\Leftrightarrow3x-1=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow3x-1=\sqrt{\left(2x-3\right)^2}=2x-3\)
\(\Leftrightarrow3x-2x=-3+1\)
\(\Leftrightarrow x=-2\)
b) Đề đúng:
\(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{3}+3}=0\)
\(\Leftrightarrow\sqrt{3-2\sqrt{2}}=\sqrt{x^2-2x\sqrt{3}+3}\)
\(\Leftrightarrow\sqrt{3-2\sqrt{2}}=\sqrt{\left(x-\sqrt{3}\right)^2}=x-\sqrt{3}\)
\(\Leftrightarrow3-2\sqrt{2}=x^2-2\sqrt{3}\cdot x+3\)
\(\Leftrightarrow-x^2+2\sqrt{3}\cdot x-2\sqrt{2}=0\)
Giải pt bậc 2 có:
\(\Delta=\left(2\sqrt{3}\right)^2-4\cdot\left(-1\right)\cdot\left(-2\sqrt{2}\right)=12-8\sqrt{2}\)
=> \(\left\{{}\begin{matrix}x_1=-\dfrac{-2\sqrt{3}+\sqrt{12-8\sqrt{2}}}{2}\\x_2=-\dfrac{-2\sqrt{3}-\sqrt{12-8\sqrt{2}}}{2}\end{matrix}\right.\)
Vậy...........................