\(\sqrt{27}-\sqrt{5\frac{1}{3}}\)\(+4,5\sqrt{2\frac{2}{3}}\)<...">
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25 tháng 8 2020

Ta có:

\(\sqrt{27}-\sqrt{5\frac{1}{3}}+4,5\sqrt{2\frac{2}{3}}+2\sqrt{27}\)

\(=3\sqrt{3}-\sqrt{\frac{16}{3}}+4,5\sqrt{\frac{8}{3}}+6\sqrt{3}\)

\(=9\sqrt{3}+\frac{4\sqrt{3}}{3}+3\sqrt{6}\)

\(=\frac{9\sqrt{6}+31\sqrt{3}}{3}\)

\(\sqrt{27}-\sqrt{5\frac{1}{3}}+4,5\sqrt{2\frac{2}{3}}+2\sqrt{27}\)

\(=\sqrt{27}-\sqrt{16.\frac{1}{3}}+4,5.\sqrt{4.\frac{1}{3}}+2\sqrt{27}\)

\(=\sqrt{27}-4\sqrt{\frac{1}{3}}+9\sqrt{\frac{1}{3}}+2\sqrt{27}\)

\(=\sqrt{27}-4\sqrt{\frac{1}{3}}+\sqrt{27}+2\sqrt{27}\)

\(=4\sqrt{27}-4\sqrt{\frac{1}{3}}\)

\(=\sqrt{54}-\sqrt{\frac{2}{3}}\)

9 tháng 10 2017

1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

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

\(=3-\sqrt{6}+2\sqrt{6}-3\)  ( vi \(\sqrt{6}-3< 0\))

\(=\sqrt{6}\)

5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)

\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)

\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)

\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)

\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)

7 tháng 8 2018

 Báo cáo sai phạm

1) 2√5−√125−√80+√605

=2√5−√52.5−√42.5+√112.5

=2√5−5√5−4√5+11√5

=4√5

2) √15−√216+√33−12√6

=√15−√62.6+√33−12√6

=√15−6√6+√33−12√6

=√(√6)2−6√6+32+√(2√6)2−12√6+32

=√(√6−3)2+√(2√6−3)2

=|√6−3|+|2√6−3|

=3−√6+2√6−3  ( vi √6−3<0)

=√6

5) 2√163 −3√127 −6√475 

=24√3 −3.13 −6√223.52 

=8√33 −1−6.25 .√13 

=8√33 −1−125 .√33 

=285 .√33 −1

24 tháng 6 2019

\(E=\)( ghi đề vào đây )

\(E=\sqrt[3]{4+\frac{5}{3}.\frac{\sqrt{31}}{\sqrt{3}}}+\sqrt[3]{4-\frac{5}{3}.\frac{\sqrt{31}}{3}}\)

\(E=\sqrt[3]{4+\frac{5\sqrt{31}}{3\sqrt{3}}}+\sqrt[3]{4+\frac{5.\sqrt{31}}{3\sqrt{3}}}\)

\(E\approx1\)

NV
24 tháng 6 2019

\(E^3=4+\frac{5}{3}\sqrt{\frac{31}{3}}+4-\frac{5}{3}\sqrt{\frac{31}{3}}+3\sqrt[3]{\left(16-\frac{25}{9}.\frac{31}{3}\right)}\left(\sqrt[3]{4+\frac{5}{3}\sqrt{\frac{31}{3}}}+\sqrt[3]{4-\frac{5}{3}\sqrt{\frac{31}{3}}}\right)\)

\(\Leftrightarrow E^3=8-7E\)

\(\Leftrightarrow E^3+7E-8=0\)

\(\Leftrightarrow\left(E-1\right)\left(E^2+E+8\right)=0\)

\(\Leftrightarrow E=1\)

Bài 1: Rút gọn biểu thức1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\)              2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)      4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) ...
Đọc tiếp

Bài 1: Rút gọn biểu thức

1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\)              2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)

3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)      4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)

5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)   6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)

7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\)  8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)

9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\)    10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)      12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

13) \(\sqrt{15-6\sqrt{6}}\)    14) \(\sqrt{8-2\sqrt{15}}\)    15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)

 

1
26 tháng 11 2017

Giúp mình :<

25 tháng 6 2019

b)\(\frac{\sqrt{27}}{\sqrt{12}}+\frac{1}{2}\)

\(=\frac{\sqrt{3}.\sqrt{9}}{\sqrt{3}.\sqrt{4}}+\frac{1}{2}\)

\(=\frac{\sqrt{9}}{\sqrt{4}}+\frac{1}{2}\)

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

\(\frac{4}{2}=2\)

25 tháng 6 2019

a) \(\sqrt{45}.\sqrt{15}.\sqrt{27}\)

\(=\left(\sqrt{15}\right)^2.\left(\sqrt{3}\right)^2.\sqrt{9}\)

\(=15.3.3\)

\(=135\)

25 tháng 6 2019

a) \(\sqrt{45}\cdot\sqrt{15}\cdot\sqrt{27}=\sqrt{45\cdot15\cdot27}=135\)

b) \(\frac{\sqrt{17}}{\sqrt{12}}+\frac{1}{2}=\frac{\sqrt{51}}{6}+\frac{3}{6}=\frac{\sqrt{51}+3}{6}\)

c) \(\sqrt{\frac{1}{3}}:\sqrt{\frac{27}{50}}\cdot\sqrt{2}=\sqrt{\frac{1}{3}\cdot\frac{50}{27}\cdot2}=\frac{10}{9}\)

d) \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}=\sqrt{9\cdot225}=45\)