\(\dfrac{8-x}{2+\sqrt[3]{x}}:\left(2+\dfrac{\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\left(\sqrt[3]{x}+\dfrac{2\sqrt[3]{x}}{\sqrt[3]{x}-2}\right).\left(\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}:\left(\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\left(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\dfrac{\sqrt[3]{x}-2}{\sqrt[3]{x}}\)

\(=\sqrt[3]{x}-2+\sqrt[3]{x}=2\sqrt[3]{x}-2\)

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

b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)

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

b, với x > 0 

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

\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)

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

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

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

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

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

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

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

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

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

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

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

\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)

a) Ta có: \(A=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)

\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}=-\sqrt{x}+1\)

b) Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=-\sqrt{6-2\sqrt{5}}+1=-\left(\sqrt{5}-1\right)+1=-\sqrt{5}+1+1=2-\sqrt{5}\)

a) ĐK: x ≥ 0; x ≠ 9; x≠4

P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

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

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

=\(\dfrac{x-4+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{x^2-4-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

=\(\dfrac{x^2-3x+2}{x-4}\)

b)  P ≤ -2

⇒ \(\dfrac{x^2-3x+2}{x-4}\) ≤ -2

⇔ \(\dfrac{x^2-3x+2}{x-4}\) + 2 ≤ 0

⇔ \(\dfrac{x^2-3x+2+2\left(x-4\right)}{x-4}\) ≤ 0

⇔ \(\dfrac{x^2-3x+2+2x-8}{x-4}\) ≤ 0

⇔\(\dfrac{x^2-x-6}{x-4}\) ≤ 0

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x-6\ge0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x-6\le0\\x-4>0\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x\le2\\3\le x< 4\end{matrix}\right.\)

Vậy.......