a: \(x=\dfrac{8}{3-\sqrt{5}}=\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\left(3+\sqrt{5}\right)=\left(\sqrt{5}+1\right)^2\)
Khi \(x=\left(\sqrt{5}+1\right)^2\) thì \(A=\dfrac{3+\sqrt{5}+3}{3+\sqrt{5}-3}=\dfrac{6+\sqrt{5}}{\sqrt{5}}\)
b: \(B=\dfrac{\sqrt{x}-7-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-7-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
c: B>-1
=>B+1>0
=>\(\dfrac{\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}>0\)
=>\(\dfrac{2\sqrt{x}-3}{\sqrt{x}-3}>0\)
=>căn x-3>0 hoặc 2căn x-3<0
=>căn x>3 hoặc căn x<3/2
=>0<=x<9/4 hoặc x>9
d: B=-4/5
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{-4}{5}\)
=>\(5\sqrt{x}=-4\sqrt{x}+12\)
=>\(9\sqrt{x}=12\)
=>\(\sqrt{x}=\dfrac{4}{3}\)
=>x=16/9
