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2 tháng 12 2017

\(=10.\dfrac{1}{10}.\dfrac{4}{3}+3.7-\dfrac{1}{6}.2\)

\(=1.\dfrac{4}{3}+21+\dfrac{1}{3}\)

\(=\dfrac{4}{3}+21+\dfrac{1}{3}\)

\(=\dfrac{28}{21}+\dfrac{441}{21}+\dfrac{7}{21}\)

\(=\dfrac{469}{21}+\dfrac{7}{21}\)

\(=\dfrac{68}{3}\)

2 tháng 12 2017

\(10.\sqrt{0,01}.\sqrt{\dfrac{16}{9}}+3.\sqrt{49}-\dfrac{1}{6}.\sqrt{4}\)

= 10 . 0,1 . \(\dfrac{4}{3}\) + 3. 7 - \(\dfrac{1}{6}.2\)

= 1 . \(\dfrac{4}{3}\) + 21 - \(\dfrac{1}{3}\)

= \(\dfrac{4}{3}+21-\dfrac{1}{3}\)

= 22

1: \(=\left(\dfrac{1}{3}\right)^{25}\cdot90^{25}\cdot\dfrac{1}{3^{25}}-\dfrac{2}{12}\)

\(=\dfrac{30^{25}}{3^{25}}-\dfrac{1}{6}=10^{25}-\dfrac{1}{6}\)

2: \(=10\cdot1\cdot\dfrac{4}{3}+7-\dfrac{1}{6}\cdot2=20\)

a: \(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\cdot12=20\)

b: \(=\dfrac{3}{5}:\left(\dfrac{-2-5}{30}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\)

\(=\dfrac{3}{5}:\dfrac{-7}{30}+\dfrac{3}{5}:\dfrac{-21}{15}\)

\(=\dfrac{3}{5}\left(\dfrac{-30}{7}-\dfrac{15}{21}\right)=\dfrac{3}{5}\cdot\left(\dfrac{-30}{7}-\dfrac{5}{7}\right)=\dfrac{3}{5}\cdot\left(-5\right)=-3\)

c: \(=5.7\left(-6.5-3.5\right)\)

\(=5.7\cdot\left(-10\right)=-57\)

d: \(=10\cdot0.1\cdot\dfrac{4}{3}+3\cdot7-\dfrac{1}{6}\cdot2\)

\(=\dfrac{4}{3}+21-\dfrac{1}{3}=22\)

21 tháng 9 2017

a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)

=1+3+5+7+9

=25

b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)

=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)

=\(\dfrac{15}{12}\)

c) =0,2+0.3+0,4

= 0.9

d) =9-8+7

=8

j) =1,2-1,3+1.4

= (-0,1)+1,4

=1,4

g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)

= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)

= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)

=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)

= \(\dfrac{71}{20}\)

Nhớ tick cho mk nha~

6 tháng 12 2017

a) \(A=\dfrac{1}{\sqrt{25}}+\dfrac{\sqrt{49}}{\sqrt{36}}-\dfrac{2}{\sqrt{100}}.\)

\(=\dfrac{1}{5}+\dfrac{7}{6}-\dfrac{1}{5}.\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\dfrac{7}{6}.\)

\(=0+\dfrac{7}{6}=\dfrac{7}{6}.\)

Vậy \(A=\dfrac{7}{6}.\)

b) \(B=\sqrt{\dfrac{0,01}{1,21}}+3.\dfrac{2}{\sqrt{10^2}+2^2+40}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+3.\dfrac{2}{10+4+40}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+3.\dfrac{1}{37}-\dfrac{3}{4}.\)

\(=\dfrac{1}{11}+\dfrac{1}{9}-\dfrac{3}{4}.\)

\(=\dfrac{36}{396}+\dfrac{44}{396}-\dfrac{297}{296}.\)

\(=-\dfrac{217}{396}.\)

Vậy \(B=-\dfrac{217}{396}.\)