a) \(\sqrt{28}+\sqrt{125}-3\sqrt{343}-\frac{3}{8}\sqrt{396}=2\sqrt{7}+5\sqrt{5}-21\sqrt{7}-\frac{9\sqrt{11}}{4}\)

\(=-19\sqrt{7}+5\sqrt{5}-\frac{9\sqrt{11}}{4}\)

b) \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

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

Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.

** Bài toán rút gọn**

Lời giải:

\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)

\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)

\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)

\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)

\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)

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\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)

\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)

\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)

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\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)

\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)

\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)

\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)

\(=\left|3-2\sqrt{2}\right|-\left|4-2\sqrt{2}\right|=3-2\sqrt{2}-4+2\sqrt{2}\)

\(=-1\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(=6\)

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

\(=2\sqrt{2}\)

c) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

\(=2\sqrt{2}\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)

`\sqrt(17-3\sqrt32)+\sqrt(17+3\sqrt32)`

`=\sqrt(17-12\sqrt2)+\sqrt(17+12\sqrt2)`

`=\sqrt(9-12\sqrt2+8)+\sqrt(9+12\sqrt2+8)`

`=\sqrt(3^2-2.3.2\sqrt2 +(2\sqrt2)^2)+\sqrt(3^2+2.3.2\sqrt2+(2\sqrt2)^2)`

`=\sqrt((3-2\sqrt2)^2)+\sqrt((3+2\sqrt2)^2)`

`=3-2\sqrt2+3+2\sqrt2`

`=6`

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

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

=6

\(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)

\(=\sqrt[3]{-64}+\sqrt{64}\)

\(=\sqrt[3]{\left(-4\right)^3}+\sqrt{8^2}\)

\(=-4+8\)

\(=4\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

\(=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{9+2.3.\sqrt{8}+8}+\sqrt{9-2.3.\sqrt{8}+8}\)

\(=\sqrt{\left(3+\sqrt{8}\right)^2}+\sqrt{\left(3-\sqrt{8}\right)^2}=\left|3+\sqrt{8}\right|+\left|3-\sqrt{8}\right|\)

\(=3+\sqrt{8}+3-\sqrt{8}\)   (do \(3>\sqrt{8}\))

\(=6\)

\(\left(\frac{2\sqrt{32}}{\sqrt{3}}-1\right):\left(7-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}}\right)\)

\(=\left[\frac{2\sqrt{96}}{3}-\frac{3}{3}\right]:\left[\frac{14}{2}-\frac{\left(\sqrt{2}-\sqrt{3}\right)\sqrt{2}}{2}\right]\)

\(=\left[\frac{8\sqrt{6}-3}{3}\right]:\left[\frac{14-2+\sqrt{6}}{2}\right]\)

\(=\left[\frac{\sqrt{3}\left(8\sqrt{2}-\sqrt{3}\right)}{3}\right]:\left[\frac{12+\sqrt{6}}{2}\right]\)

\(=\left[\frac{\sqrt{3}\left(8\sqrt{2}-\sqrt{3}\right)}{3}\right]:\left[\frac{\sqrt{3}\left(2\sqrt{6}+1\right)}{\sqrt{2}}\right]\)

\(=\frac{8\sqrt{2}-\sqrt{3}}{\sqrt{3}}.\frac{\sqrt{2}}{\sqrt{3}\left(2\sqrt{6}+1\right)}\)

\(=\frac{16-\sqrt{6}}{6\sqrt{6}+3}\)

\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)

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