Bài 3:
1. \(\sqrt{x^2-4x+4}=\left(x-2\right)^5\left(x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\left(x-2\right)^5\)
\(\Leftrightarrow\left|x-2\right|=\left(x-2\right)^5\)
\(\Leftrightarrow\left(x-2\right)-\left(x-2\right)^5=0\)
\(\Leftrightarrow\left(x-2\right)\left[1-\left(x-2\right)^4\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\1-\left(x-2\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x-2\right)^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
2. \(M=\dfrac{4}{\sqrt{4+2\sqrt{3}}}+\dfrac{1}{\sqrt{3}-2}-\dfrac{6}{\sqrt{12-6\sqrt{3}}}\)
\(M=\dfrac{4}{\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}+\dfrac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{6}{\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)
\(M=\dfrac{4}{\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{3}+2}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{6}{\sqrt{\left(3-\sqrt{3}\right)^2}}\)
\(M=\dfrac{4}{\sqrt{3}+1}+\dfrac{\sqrt{3}+2}{3-4}-\dfrac{6}{3-\sqrt{3}}\)
\(M=\dfrac{4\left(\sqrt{3}-1\right)}{3-1}+\dfrac{\sqrt{3}+2}{-1}-\dfrac{6\left(3+\sqrt{3}\right)}{9-3}\)
\(M=\dfrac{4\left(\sqrt{3}-1\right)}{2}-\left(\sqrt{3}+2\right)-\dfrac{6\left(3+\sqrt{3}\right)}{6}\)
\(M=2\sqrt{3}-2-\sqrt{3}-2-3-\sqrt{3}\)
\(M=-7\)
