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21 tháng 10 2018

\(\sqrt{81.16.169}=\sqrt{81}.\sqrt{16}.\sqrt{169}=9.4.13=468\)

\(\sqrt{10}.\sqrt{810}=\sqrt{10.10}.\sqrt{81}=10.9=90\)

\(\sqrt{64}.\sqrt{81.100}-\sqrt{64}.\sqrt{196.16}=\sqrt{64}\left(\sqrt{81}.\sqrt{100}-\sqrt{196}.\sqrt{16}\right)=8.\left(9.10-14.4\right)=8.34=272\)

21 tháng 10 2018

Hông có gì :)))

14 tháng 5 2021

a) \(\dfrac{40}{27}\)

b) \(\dfrac{196}{45}\)

c) \(\dfrac{56}{9}\)

d) 1296

19 tháng 5 2021

a) \sqrt{\dfrac{25}{81} \cdot \dfrac{16}{49} \cdot \dfrac{196}{9}}

=\sqrt{\dfrac{25}{81}} \cdot \sqrt{\dfrac{16}{49}} \cdot \sqrt{\dfrac{196}{9}}

=\sqrt{\left(\dfrac{5}{9}\right)^{2}} \cdot \sqrt{\left(\dfrac{4}{7}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{3}\right)^{2}}

=\dfrac{5}{9} \cdot \dfrac{4}{7} \cdot \dfrac{14}{3}=\dfrac{40}{27}.

b) \sqrt{3 \dfrac{1}{16} \cdot 2 \dfrac{14}{25} \cdot 2 \dfrac{34}{81}}

=\sqrt{\dfrac{49}{16} \cdot \dfrac{64}{25} \cdot \dfrac{196}{81}}

=\sqrt{\dfrac{49}{16}} \cdot \sqrt{\dfrac{64}{25}} \cdot \sqrt{\dfrac{196}{81}}

=\sqrt{\left(\dfrac{7}{4}\right)^{2}} \cdot \sqrt{\left(\dfrac{8}{5}\right)^{2}} \cdot \sqrt{\left(\dfrac{14}{9}\right)^{2}}

=\dfrac{7}{4} \cdot \dfrac{8}{5} \cdot \dfrac{14}{9}=\dfrac{196}{45}.

c) \dfrac{\sqrt{640} \cdot \sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.343}{567}}

=\sqrt{\dfrac{64.49 .7}{81.7}}=\sqrt{\dfrac{64.49}{81}}

=\dfrac{\sqrt{64} \cdot \sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}

=\dfrac{56}{9}.

d) \sqrt{21,6} \cdot \sqrt{810} \cdot \sqrt{11^{2}-5^{2}}

=\sqrt{21,6.810 \cdot\left(11^{2}-5^{2}\right)}

=\sqrt{216.81 .(11+5)(11-5)}

=\sqrt{36.6 .9^{2} \cdot 4^{2} .6}

=\sqrt{36^{2} .9^{2} \cdot 4^{2}}=36.9 .4=1296.

17 tháng 7 2021

đó là số 2 ko phải chữ s mik xin lỗi

Ta có: \(\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{2-2\sqrt{2}\cdot1+1}}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{\left(\sqrt{2}-1\right)^2}}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\left|\sqrt{2}-1\right|}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\left(\sqrt{2}-1\right)}}\)(Vì \(\sqrt{2}>1\))

\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{2}+8}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{18-8\sqrt{2}}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}\)

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

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

\(=\sqrt{4\sqrt{2}+4\left(4-\sqrt{2}\right)}\)(Vì \(4>\sqrt{2}\))

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

\(=\sqrt{16}=4\)

Sửa đề: \(\sqrt[3]{24}-4\sqrt[3]{\dfrac{1}{9}}+3\sqrt[3]{81}=3\sqrt{3}\cdot x\)

\(\Leftrightarrow x\cdot3\sqrt{3}=2\sqrt[3]{3}-\dfrac{4}{3}\sqrt[3]{3}+9\sqrt[3]{3}\)

\(\Leftrightarrow x\cdot3\sqrt{3}=\dfrac{29}{3}\sqrt[3]{3}\)

hay \(x=\dfrac{29}{3}\sqrt[3]{3}:3\sqrt{3}=\dfrac{\dfrac{29}{9}\sqrt[3]{3}}{\sqrt{3}}\)

15 tháng 7 2016

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

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

\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)

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

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

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

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

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

15 tháng 7 2016

\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)

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

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

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

\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)

\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)

7 tháng 8 2017

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

\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)

\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)

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

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

\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)

https://hoc24.vn/hoi-dap/question/405366.html

\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

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

= 5 - 3

= 2

13 tháng 3 2020
https://i.imgur.com/LeR5GY4.jpg

a: \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

b: \(=\dfrac{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)

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

\(=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)

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

\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)

hay \(B=2\sqrt{10}\)

d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)

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

hay \(D=\sqrt{2}\)