Câu 1:
a) Ta có: \(A=\sqrt{18}-3\sqrt{50}+\dfrac{5}{6}\sqrt{72}\)
\(=3\sqrt{2}-3\cdot5\sqrt{2}+\dfrac{5}{6}\cdot6\sqrt{2}\)
\(=-7\sqrt{2}\)
b) Ta có: \(B=\sqrt{6-2\sqrt{5}}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\sqrt{5}-1-\sqrt{5}+2\)
=1
c) Ta có: \(C=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}+2\right)\)
\(=\sqrt{3}+2+\sqrt{2}+1-\sqrt{3}-\sqrt{2}-2\)
=1
Câu 2:
a) Ta có: \(\sqrt{x^2-4x+4}=5\)
\(\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) Ta có: \(3\sqrt{4x-8}-\dfrac{2}{5}\sqrt{25x-50}+\sqrt{9x-18}=7\)
\(\Leftrightarrow6\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=7\)
\(\Leftrightarrow x-2=1\)
hay x=3
