1: \(A=\sqrt{21-4\sqrt{5}}-\dfrac{\sqrt{160}}{\sqrt{8}}\)
\(=\sqrt{20-2\cdot2\sqrt{5}\cdot1+1}-\sqrt{20}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}-2\sqrt{5}\)
\(=2\sqrt{5}-1-2\sqrt{5}=-1\)
2:
a: \(B=\left(\dfrac{x+2\sqrt{x}+4}{\sqrt{x^3}-8}+\dfrac{x+2\sqrt{x}+1}{x-1}\right):\dfrac{3x-9}{x-\sqrt{x}-2}\)
\(=\left(\dfrac{x+2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}+\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
\(=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
\(=\dfrac{\sqrt{x}-1+x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)
\(=\dfrac{x-3}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{3\left(x-3\right)}=\dfrac{\sqrt{x}+1}{3\left(\sqrt{x}-1\right)}\)
b: Để B>0 thì \(\dfrac{\sqrt{x}+1}{3\left(\sqrt{x}-1\right)}>0\)
=>\(\sqrt{x}-1>0\)
=>x>1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>1\\x\notin\left\{3;4\right\}\end{matrix}\right.\)
