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2 tháng 8 2017

B=(1-2/5) (1-2/7) (1-2/9) ...... (1-2/99)

  =3/5.5/7.7/9 .....97/99

  =3/99=1/33

8 tháng 8 2017

B=(1-2/5)(1-2/7)(1-2/9)......(1-2/99)

=3/5.5/7.7/9........97/99

=3/5.5/7.7/9.....97/99

=3/99

=1/33

20 tháng 6 2018

= (11+9+2)+(1/2-3/2+5/2)-(2/3+5/3-7/3)

=22+1,5-0

=23.5

24 tháng 2 2023

=(11+9+2)+(1/2-3/2+5/2)-(2/3+5/3-7/3)

=22+1,5-0

=23.5

Chúc học tốt

18 tháng 9 2016

Tìm x . biết : 

\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

2 tháng 3 2020

a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)

\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)

\(\Rightarrow-x=1\)

\(\Rightarrow x=-1\)

10 tháng 6 2018

Bài 2:

\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{101}\)

\(\Leftrightarrow A=\dfrac{100}{101}\)

Vậy ...

\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}\)

\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)

\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{16}\right)\)

\(\Leftrightarrow B=\dfrac{1}{3}.\dfrac{15}{16}\)

\(\Leftrightarrow B=\dfrac{5}{16}\)

Vậy ...

10 tháng 6 2018

Bài 1:

B=\(\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}\right)}\)

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

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

\(=\dfrac{\left(2^3+2\right)\left(2+1\right)+1}{2^4}.\dfrac{2^4}{\left(2^3+2\right)\left(2-1\right)}\)

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

\(=\dfrac{2.5.3+1}{2.5.1+1}\)

\(=\dfrac{31}{11}\)

\(=2,\left(81\right)\)

16 tháng 8 2017

a) (x-3).11=(x-7).10

=>11x-33=10x-70

=>11x-10x=-70+33

=>x= -37

28 tháng 8 2018

C =\(\frac{1}{100}-\frac{1}{100.99}-...\)\(-\frac{1}{3.2}-\frac{1}{2.1}\)

C = \(\frac{1}{100}-\frac{1}{100}+\frac{1}{99}-\frac{1}{99}+...\)\(+\frac{1}{3}-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)

C = 1