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29 tháng 10 2021

\(K=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}-\sqrt{\left(-8\right)^2}\)

    \(=\sqrt{81-17}-8=\sqrt{64}-8=8-8=0\)

29 tháng 10 2021

\(=\sqrt{81-17}-8\)

=8-8

=0

1 tháng 11 2020

a) \(H=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)

\(=\sqrt{81-17}=\sqrt{64}=8\)

b) \(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right).\sqrt{5}\)

\(=\sqrt{20}.\sqrt{5}-3\sqrt{5}.\sqrt{5}+\sqrt{80}.\sqrt{5}\)

\(=\sqrt{100}-3.5+\sqrt{400}=\sqrt{10^2}-15+\sqrt{20^2}\)

\(=10-15+20=15\)

1 tháng 11 2020

\(H=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)   

\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)   

\(=\sqrt{9^2-\left(\sqrt{17}\right)^2}\)   

\(=\sqrt{81-17}\)   

\(=\sqrt{64}=8\)   

\(K=\left(\sqrt{20}-3\sqrt{5}+\sqrt{80}\right)\cdot\sqrt{5}\)   

\(=\sqrt{20}\cdot\sqrt{5}-3\sqrt{5}\cdot\sqrt{5}+\sqrt{80}\cdot\sqrt{5}\)   

\(=\sqrt{20\cdot5}-3\sqrt{5\cdot5}+\sqrt{80\cdot5}\)   

\(=\sqrt{100}-3\sqrt{25}+\sqrt{400}\)   

\(=10-3\cdot5+20\)   

\(=15\)

a) Ta có: \(VT=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)

\(=\sqrt{\left(9-\sqrt{17}\right)\cdot\left(9+\sqrt{17}\right)}\)

\(=\sqrt{81-17}=\sqrt{64}=8\)=VP(đpcm)

b) Ta có: \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)

=9=VP(đpcm)

26 tháng 6 2019

\(a,\sqrt{\frac{5.\left(38^2-17^2\right)}{8.\left(47^2-19^2\right)}}\)

\(=\sqrt{\frac{5.\left(38-17\right)\left(38+17\right)}{8.\left(47-19\right)\left(47+19\right)}}\)

\(=\sqrt{\frac{5.21.55}{8.28.66}}\)

\(=\sqrt{\frac{5775}{14784}}=\frac{5\sqrt{231}}{2\sqrt{4370}}\)

26 tháng 6 2019

.bn tính lại \(\sqrt{14784}\)đi sao lạ vậy

AH
Akai Haruma
Giáo viên
28 tháng 5 2019

Bài 2:

a)

\(\sqrt{9-\sqrt{17}}-\sqrt{9+\sqrt{17}}=\sqrt{\frac{18-2\sqrt{17}}{2}}-\sqrt{\frac{18+2\sqrt{17}}{2}}\)

\(=\sqrt{\frac{17+1-2\sqrt{17}}{2}}-\sqrt{\frac{17+1+2\sqrt{17}}{2}}=\sqrt{\frac{(\sqrt{17}-1)^2}{2}}-\sqrt{\frac{(\sqrt{17}+1)^2}{2}}\)

\(=\frac{\sqrt{17}-1}{\sqrt{2}}-\frac{\sqrt{17}+1}{\sqrt{2}}=-\sqrt{2}\)

b)

\(2\sqrt{2}(\sqrt{3}-2)+(1+2\sqrt{2})^2-2\sqrt{6}\)

\(=2\sqrt{6}-4\sqrt{2}+(1+4\sqrt{2}+8)-2\sqrt{6}\)

\(=1+8=9\)

AH
Akai Haruma
Giáo viên
28 tháng 5 2019

Bài 1:

a)

\(\frac{\sqrt{6}+\sqrt{16}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+4}{2(\sqrt{3}+\sqrt{7})}=\frac{1}{2}.\frac{(\sqrt{6}+4)(\sqrt{7}-\sqrt{3})}{(\sqrt{3}+\sqrt{7})(\sqrt{7}-\sqrt{3})}\)

\(=\frac{(4+\sqrt{6})(\sqrt{7}-\sqrt{3})}{8}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{16}-\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+1)(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)

11 tháng 7 2017

a) \(VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)

=\(\sqrt{9^2-\left(\sqrt{17}\right)^2}=\sqrt{81-17}=\sqrt{64}=8=VP\)

b) \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

=\(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9=VP\)

NV
25 tháng 6 2021

\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)

\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)

\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)

\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)

8 tháng 7 2015

thì rút gọn vế trái là ra