\(1,\\ A=\left(2\sqrt{6}-4\sqrt{3}-\sqrt{6}\right)\cdot\sqrt{6}+12\sqrt{2}\\ A=6-12\sqrt{2}+12\sqrt{2}=6\\ B=\left(\dfrac{\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}+\sqrt{5}\right):2\sqrt{5}\\ B=\dfrac{1}{10}-\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{1}{5}\\ C=\sqrt{21+12\sqrt{3}}-\sqrt{21-12\sqrt{3}}\\ C=\sqrt{\left(2\sqrt{3}+3\right)^2}-\sqrt{\left(2\sqrt{3}-3\right)^2}\\ C=2\sqrt{3}+3-2\sqrt{3}+3=6\)
\(2,\\ a,A=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,x=7-2\sqrt{6}\Leftrightarrow\sqrt{x}=\sqrt{6}-1\\ \Leftrightarrow A=\dfrac{\sqrt{6}-1-1}{\sqrt{6}+1-1}=\dfrac{\sqrt{6}-2}{\sqrt{6}}=\dfrac{6-2\sqrt{6}}{6}=\dfrac{3-\sqrt{6}}{3}\\ A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\\ Ta.có.\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\\ \Leftrightarrow1-\dfrac{2}{\sqrt{x}-1}\ge1-2=-1\\ A_{min}=-1\Leftrightarrow x=0\)
\(3,\\ a,ĐK:x\ge\dfrac{1}{3}\\ PT\Leftrightarrow6x-2=16\Leftrightarrow x=3\left(tm\right)\\ b,ĐK:x\ge2\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{1}{9}\sqrt{x-2}=-4\\ \Leftrightarrow-\sqrt{x-2}=-4\\ \Leftrightarrow\sqrt{x-2}=4\Leftrightarrow x-2=16\Leftrightarrow x=18\left(tm\right)\\ c,ĐK:x\ge0\\ PT\Leftrightarrow\left|3x+2\right|=4x\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x\ge-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\KTMĐKXĐ\end{matrix}\right.\)
\(d,ĐK:x\ge1\\ PT\Leftrightarrow x-2\sqrt{x-1}=x-1\\ \Leftrightarrow2\sqrt{x-1}=1\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\\ \Leftrightarrow x-1=\dfrac{1}{4}\Leftrightarrow x=\dfrac{5}{4}\)
PP