\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)

\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)

\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)

\(=11.2.13.\sqrt{9}-1=286.3-1=857\)

\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)

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

\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)

\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)

giúp nốt mình câu e đi bạn !

\(a=26\sqrt{3}\)

Bạn làm câu b luôn dùm mk với

`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.

`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.

`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.

`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.

`= 7`.

`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`

`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`

`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`

`= 33 - 11 = 22`.

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

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

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)

  \(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)

  \(=0\)

b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)

Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)

\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)

Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

a) E=0

b) F=căn 3/2

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

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

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

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

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