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

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

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

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

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

1:

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

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)

Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)

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

\(=\sqrt{2}\)

\(A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}-\sqrt{37-20\sqrt{3}}\)

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

\(=2+\sqrt{3}+2-\sqrt{3}-5+2\sqrt{3}\)

\(=2\sqrt{3}-1\)

\(=\dfrac{\left(\sqrt{1}-\sqrt{2}\right)}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+....+\dfrac{\sqrt{899}-\sqrt{900}}{899-900}\\ =\sqrt{900}-\sqrt{899}+\sqrt{899}-.....+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}=\\ =\sqrt{900}-\sqrt{1}\\ =30-1=29\)

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

\(=\sqrt{2}+\dfrac{\sqrt{3}+1}{2}=\dfrac{2\sqrt{2}+\sqrt{3}+1}{2}\)

Đề lỗi không?

Chắc đề là: \(\dfrac{\sqrt{2}}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{4}-\sqrt{\dfrac{6+2\sqrt{5}}{4}}\)

\(=\dfrac{\sqrt{10}-\sqrt{2}}{4}-\sqrt{\left(\dfrac{\sqrt{5}+1}{2}\right)^2}=\dfrac{\sqrt{10}-\sqrt{2}}{4}-\dfrac{\sqrt{5}+1}{2}=\dfrac{\sqrt{10}-\sqrt{2}-2\sqrt{5}-2}{4}\)

Ta có: M=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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