-20/147

=\(\frac{-4}{15}\)

\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\cdot\left(\frac{-4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right)\cdot\frac{5}{7}}\)

\(=-\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right)\cdot\frac{4}{15}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right)\cdot\frac{5}{7}}\)

\(=-\frac{\frac{4}{15}\cdot\frac{5-4\sqrt{2}}{70}}{\frac{5}{7}\cdot\frac{5-4\sqrt{2}}{50}}\)

\(=-\frac{4\left(5-4\sqrt{2}\right)}{15\left(5-4\sqrt{2}\right)}\)

\(=-\frac{4}{15}\)

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

A=\(\sqrt{\frac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.3\sqrt{2}+\sqrt{14}=\sqrt{\frac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\sqrt{2}\left(3+\sqrt{7}\right)\)

\(8>3.\sqrt{7}\Rightarrow8-3\sqrt{7}>0\left(lienhop\right)\left(8-3\sqrt{7}\right)\)

\(A=\sqrt{\left(8-3.\sqrt{7}\right)}.\sqrt{2}\left(3+\sqrt{7}\right)\)

\(A=\sqrt{\left(16-2.3.\sqrt{7}\right)}.\left(3+\sqrt{7}\right)\)

\(A=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}.\left(3+\sqrt{7}\right)\)

\(A=\sqrt{\left(3-\sqrt{7}\right)^2}\left(3+\sqrt{7}\right)\)

\(3-\sqrt{7}>0\)

\(\Rightarrow A=\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)=9-7=2\)

\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}\cdot\sqrt{2}\left(3+\sqrt{7}\right)\\ =\sqrt{\dfrac{2\left(3+\sqrt{7}\right)^2}{8+3\sqrt{7}}}=\sqrt{\dfrac{32+12\sqrt{7}}{8+3\sqrt{7}}}\\ =\sqrt{\dfrac{4\left(8+3\sqrt{7}\right)}{8+3\sqrt{7}}}=\sqrt{4}=2\)

\(A=\sqrt{\frac{1}{8+3\sqrt{7}}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(A=\sqrt{\frac{2}{16+6\sqrt{7}}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(A=\frac{\sqrt{2}}{3+\sqrt{7}}\left(3+\sqrt{7}\right)\sqrt{2}\)
\(A=2\)

tu lam di cau nao kho thi hoi hoi vay ko ai tra loi cho dau

cau e)

\(A=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)(suy ra A>=0)

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

\(A^2=1\)

A=1

(bai toan co nhieu cach)

cau m)

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

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

\(=1\)

cau G)

\(=\frac{5\sqrt{7}}{\sqrt{35}}-\frac{7\sqrt{5}}{\sqrt{35}}+\frac{2\sqrt{70}}{\sqrt{35}}\)

\(=\frac{5}{\sqrt{5}}-\frac{7}{\sqrt{7}}+2\sqrt{2}\)

\(=\sqrt{5}-\sqrt{7}+2\sqrt{2}\)

1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)

\(=7-2\sqrt{4\sqrt{7}}\)

cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với