Câu 1:
\(a,=10\sqrt{5}-6\sqrt{3}+\sqrt{5}-6\sqrt{3}=11\sqrt{5}-12\sqrt{3}\\ b,=\dfrac{9\left(\sqrt{10}+1\right)}{9}+\dfrac{\sqrt{5}\left(\sqrt{10}-1\right)}{\sqrt{5}}=\sqrt{10}+1+\sqrt{10}-1=2\sqrt{10}\)
Câu 2:
\(a,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\\ b,\Leftrightarrow\sqrt{\left(3x-1\right)^2}=9\\ \Leftrightarrow\left|3x-1\right|=9\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=9\\1-3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\\ c,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x-1}=2\sqrt{x}-1\\ \Leftrightarrow2x-1=4x-4\sqrt{x}+1\\ \Leftrightarrow2x-4\sqrt{x}+2=0\\ \Leftrightarrow2\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\left(tm\right)\)
