1)\(\sqrt{4x^2-9}=2\sqrt{2x+3}\)(Đkxđ \(x\ge\dfrac{3}{2}\))
\(\Leftrightarrow\sqrt{\left(2x-3\right)\left(2x+3\right)}=2\sqrt{2x+3}\)
\(\Leftrightarrow\sqrt{2x-3}=2\)
\(\Leftrightarrow2x-3=4\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(N\right)\)
2)\(\sqrt{3x^2-4x+3}=1-2x\) Đkxđ:\(3x^2-4x+3=\left(x\sqrt{3}-\dfrac{2}{\sqrt{3}}\right)^2+\dfrac{5}{3}>0\)
\(\Leftrightarrow3x^2-4x+3=\left(1-2x\right)^2\)
\(\Leftrightarrow3x^2-4x+3=1-4x+4x^2\)
\(\Leftrightarrow x^2-2=0\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)
1: Ta có: \(\sqrt{4x^2-9}=2\sqrt{2x+3}\)
\(\Leftrightarrow4x^2-9=8x+12\)
\(\Leftrightarrow4x^2-8x-21=0\)
\(\Leftrightarrow4x^2-8x+4-25=0\)
\(\Leftrightarrow\left(2x-2-5\right)\left(2x-2+5\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\left(nhận\right)\\x=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
2: Ta có: \(\sqrt{3x^2-4x+3}=1-2x\)
\(\Leftrightarrow3x^2-4x+3=4x^2-4x+1\)
\(\Leftrightarrow-x^2=-2\)
\(\Leftrightarrow x^2=2\)
hay \(x=-\sqrt{2}\)
3: Ta có: \(\sqrt{x^2-4}-x+2=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x-2\)
\(\Leftrightarrow x^2-4=x^2-4x+4\)
\(\Leftrightarrow-4x=-8\)
hay \(x=2\left(nhận\right)\)
4: Ta có: \(\sqrt{x^2-2x+1}=x^2-1\)
\(\Leftrightarrow x^2-2x+1=\left(x-1\right)^2\cdot\left(x+1\right)^2\)
\(\Leftrightarrow\left(x-1\right)^2\left[1-\left(x+1\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(1-x-1\right)\left(1+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot x\cdot\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=0\left(loại\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
5: Ta có: \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)
\(\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3-\sqrt{2}\left(x\ge-\dfrac{1}{3}\right)\\3x+1=\sqrt{2}-2\left(x< -\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=2-\sqrt{2}\\3x=\sqrt{2}-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-\sqrt{2}}{3}\left(nhận\right)\\x=\dfrac{\sqrt{2}-3}{3}\left(nhận\right)\end{matrix}\right.\)
