\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)

\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)

\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)

\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x+\sqrt{x}-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Thay \(x=6-2\sqrt{5}\) vào B ta có:

\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{6-2\sqrt{5}}-1}{\sqrt{6-2\sqrt{5}}+1}\\ =\dfrac{\sqrt{5-2\sqrt{5}+1}-1}{\sqrt{5-2\sqrt{5}+1}+1}\\ =\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\\ =\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}\\ =\dfrac{\sqrt{5}-2}{\sqrt{5}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{5}\\ =\dfrac{5-2\sqrt{5}}{5}\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

\(A=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\left(x>0,x\ne1\right)\)

\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)

\(=\dfrac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\)

\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4;9\right)\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(C=\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\left(\dfrac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)

\(=\dfrac{x+\sqrt{x}-1-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\dfrac{3\sqrt{x}-2}{x+\sqrt{x}+1}\)

Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) −  \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)

                                           =2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\) 

                                           =(2− 3+ 8)\(\sqrt{2}\)

                                           =7\(\sqrt{2}\)

Câu 2: Mik ko chắc làm đúng hay ko  nên ko làm

b: \(=\dfrac{\sqrt{x}-4+\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-4}\)

`#Hưng`

\(a,3\sqrt{8\sqrt{5}}-2\sqrt{9\sqrt{20}}\\ =\sqrt{9.8\sqrt{5}}-\sqrt{4.9\sqrt{20}}\\ =\sqrt{72\sqrt{5}}-\sqrt{36\sqrt{20}}\\ =\sqrt{\sqrt{5184.5}}-\sqrt{\sqrt{1296.20}}\\ =\sqrt{\sqrt{25920}}-\sqrt{\sqrt{25920}}\\ =0\)

\(b,ĐKXĐ:x\sqrt{x}-\sqrt{x}+x-1\ne0\\ \Rightarrow\sqrt{x}\left(x-1\right)+\left(x-1\right)\ne0\\ \Rightarrow\left(x-1\right)\left(\sqrt{x}+1\right)\ne0\\ \Rightarrow x-1\ne0\left(vì.\sqrt{x}+1>0\right)\\ \Rightarrow x\ne1\)

a: ĐKXĐ: x>0

\(E=\dfrac{\sqrt{x}}{x+2\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)

\(=\dfrac{1}{\sqrt{x}+2}:\dfrac{\sqrt{x}+2+x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\dfrac{1}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

b: E=2/5

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{2}{5}\)

=>\(5\sqrt{x}=2x+2\sqrt{x}+4\)

=>\(2x-3\sqrt{x}+4=0\)

=>\(x-\dfrac{3}{2}\cdot\sqrt{x}+2=0\)

=>\(x-2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)

=>\(\left(\sqrt{x}-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)(vô lý)

Vậy: \(x\in\varnothing\)