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19 tháng 6 2019

1.

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

2.

\(\sqrt{\frac{14+5\sqrt{3}}{2}}\\ =\frac{\sqrt{14+5\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{28+10\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(5+\sqrt{3}\right)^2}}{2}\\ =\frac{5+\sqrt{3}}{2}\)

19 tháng 6 2019

Hỏi đáp Toán

a) Ta có:

5√15+12√20+√5515+1220+5

=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35

b)  Ta có: 

√12+√4,5+√12,512+4,5+12,5

=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922

c) Ta có:

√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5

d) Ta có:

0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2



 

Bạn giải bài đâu vậy? Kiếm điểm hỏi đáp hở, Boy anime?

19 tháng 6 2019

1/\(\sqrt{\frac{4}{5}}\)+\(\sqrt{\frac{1}{2}}\)

=\(\sqrt{\frac{4.5}{5.5}}\)+\(\sqrt{\frac{1.2}{2.2}}\)

= \(5.2.\sqrt{5}\)+\(2\sqrt{2}\)

=\(10\sqrt{5}+2\sqrt{2}\)

19 tháng 6 2019

2.

\(\sqrt{\frac{1}{12}}\)+\(\sqrt{\frac{1}{3}}\)

=\(\sqrt{\frac{1.12}{12.12}}\)+\(\sqrt{\frac{1.3}{3.3}}\)

=\(12.2\sqrt{3}\)+\(3\sqrt{3}\)

=\(\sqrt{3}\left(24+3\right)\)

=\(27\sqrt{3}\)

6 tháng 7 2019

EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.

1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)

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

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

3) \(\sqrt{\frac{1}{a\left(1-\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-3\right)}}=\sqrt{-\frac{1+\sqrt{3}}{2a}}\)

4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)

22 tháng 8 2020

a, \(\sqrt{\frac{1}{60}}=\frac{\sqrt{1}}{\sqrt{60}}=\frac{\sqrt{1}.\sqrt{60}}{\sqrt{60}.\sqrt{60}}=\frac{\sqrt{60}}{60}=\frac{2.\sqrt{15}}{2.30}=\frac{\sqrt{15}}{30}\)

c, \(\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\)

d, \(\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{\left(\sqrt{7}-\sqrt{3}\right)^2}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}=\frac{7-2\sqrt{21}+3}{7-3}=\frac{10-2\sqrt{21}}{4}\)

17 tháng 5 2021

\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

17 tháng 5 2021

\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)