\(\frac{1}{2}\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\...">
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12 tháng 4 2017

1/2 . ( 2/9 + 3/7 - 5/27 )

=1/2 . ( 2/9-5/27)+3/7

=1/2 . (1/27+3/7)

=1/2 .

12 tháng 4 2017

=1/2x88/189=44/189

25 tháng 5 2020

a)(1/4+3/4)-(5/13+8/13)+2/11=1-1+2/11=2/11

b)(21/31+10/31)-16/7+(44/53+9/53)=1+1-16/7=14/7-16/7=-2/7

bằng2/11

         -16/7

12 tháng 2 2017

a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)

= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)

b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)

\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)

\(=66+44+33+22+12=177\)

c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)

= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)

= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)

= còn lại làm nốt nha! bận ròy

12 tháng 2 2017

gidkjbibvvfrxdrfdfsddf

25 tháng 5 2020

\(\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)

\(=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)

\(=-1+1+\frac{-1}{5}\)

\(=-\frac{1}{5}\)

25 tháng 5 2020

=(-5/7-2/7)+(3/4+1/4)-1/5=-1+1-1/5=-1/5

(-5/7+-2/7)+(3/4+1/4)+-1/5

-1+1+-1/5

-1/5

14 tháng 4 2019

\(2.THPT\)

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9.\frac{99}{100}\)

\(A=\frac{891}{100}\)

\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)

\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)

\(B=\frac{1}{5}-\frac{1}{95}\)

\(B=\frac{18}{95}\)

\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(D=\frac{1}{2}-\frac{1}{28}\)

\(D=\frac{13}{28}\)

9 tháng 7 2021

1. 

a.\(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

b. \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\)

c. \(\left(\frac{-3}{5}\right)^5=\frac{-243}{3125}\)

d. \(\left(\frac{-1}{5}\right)^2=\frac{1}{25}\)

e. \(\left(\frac{-1}{6}\right)^3=\frac{-1}{216}\)

10 tháng 7 2021

Trả lời:

Bài 1: 

a, \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{16}\)

b, \(\left(\frac{1}{2}\right)^3=\frac{1^3}{2^3}=\frac{1}{8}\)

c, \(\left(\frac{-3}{5}\right)^2=\frac{\left(-3\right)^2}{5^2}=\frac{9}{25}\)

d, \(\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)

e, \(\left(\frac{-1}{6}\right)^3=\frac{\left(-1\right)^3}{6^3}=\frac{-1}{216}\)

Bài 2:

a, \(\left(\frac{3}{2}\right)^2.\left(\frac{4}{3}\right)^2=\frac{9}{4}.\frac{16}{9}=4\)

b, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)

c, \(\left(-\frac{1}{2}\right)^2.\left(\frac{2}{5}\right)^2=\frac{1}{4}.\frac{4}{25}=\frac{1}{25}\)

d, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)

e, \(\left(-5\right)^3.\frac{1}{5}=-125.\frac{1}{5}=-25\)

f, \(\left(\frac{2}{9}\right)^5.\left(-\frac{27}{4}\right)^5=\frac{2^5}{9^5}.\frac{\left(-27\right)^5}{4^5}=\frac{2^5.\left(-27\right)^5}{9^5.4^5}=\frac{2^5.\left[\left(-3\right)^3\right]^5}{\left(3^2\right)^5.\left(2^2\right)^5}=-\frac{2^5.3^{15}}{3^{10}.2^{10}}=\frac{3^5}{2^5}\)