b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{12-3\sqrt{7}}-\sqrt{2}\cdot\sqrt{12+3\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{21}\right)^2-2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}\right)^2+2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)

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

\(=-\sqrt{6}\)  

c) \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)

\(=\sqrt[3]{\dfrac{3\cdot9}{4\cdot16}}\)

\(=\sqrt[3]{\left(\dfrac{3}{4}\right)^3}\)

\(=\dfrac{3}{4}\)

d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)

\(=\sqrt[3]{\dfrac{54}{-2}}\)

\(=\sqrt[3]{-27}\)

\(=\sqrt[3]{\left(-3\right)^3}\)

\(=-3\) 

a: Sửa đề: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}\cdot\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{2\sqrt{2}\left(\sqrt{6}+1\right)+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)

e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{2\sqrt{2}+3\sqrt{2}+6+1}-\sqrt[3]{2\sqrt{2}-3\sqrt{2}+6-1}\)

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

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

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

đừng để ý cái đen đen🥲

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy

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

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

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

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

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

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

giúp mình với ạ 

b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)

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

\(=2\sqrt{3}-4\)

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

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

\(=\sqrt{2}\)

Đề lỗi không?