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14 tháng 5 2019

Mk giải ko chép lại đề nhá!

Bài 3: 

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}\)\(-\frac{1}{50}\)

\(=\frac{1}{1}-\frac{1}{50}\)

\(=\frac{50}{50}-\frac{1}{50}\)

\(=\frac{49}{50}\)

Vậy: M < 1

14 tháng 5 2019

Bài 2:

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(=\frac{1}{1}-\frac{1}{2015}\)

\(=\frac{2015}{2015}-\frac{1}{2015}\)

\(=\frac{2014}{2015}\)

25 tháng 5 2016

A=1+2+3+4+5+...+99+100

A=(1+100).100:2=101.50=5050

B=1/2+1/6+1/12+1/20+1/30+...+1/9900

B=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+....+1/99.100

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100

B=1-1/100=99/100

25 tháng 5 2016

A = 100 x 101 : 2 = 5050

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)

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

      \(=1-\frac{1}{100}\)

        \(=\frac{99}{100}\)

30 tháng 4 2019

1/ Tính:

\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\) 

\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\) 

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\) 

\(=1-\frac{1}{10}\) 

\(=\frac{9}{10}\)

10 tháng 7 2019

Bạn gõ lại đề đi :v

Đọc chả hiểu đề gì cả ... đề k có x

Mà phía dưới có cái đáp số x= ... là sao ??

10 tháng 7 2019

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

5 tháng 8 2017

\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{4}{8}-\frac{1}{8}\)

\(=\frac{3}{8}\)

\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)

\(=1-0-0-0-...-0-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\frac{50}{51}\)

\(=\frac{25}{17}\)

\(d,\)giống câu a tự làm nha mỏi tay quá.

5 tháng 8 2017

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)

=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)

=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)

4 tháng 5 2016

 nhung ma ko cothoi gian giai

4 tháng 5 2016

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

DD
18 tháng 5 2021

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

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

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

\(A=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(B=1-\frac{1}{101}=\frac{100}{101}\)

DD
18 tháng 5 2021

\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)

\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)

\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)

\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)

\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)

18 tháng 2 2017

a ) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{4}{20}+\frac{3}{20}\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\left(\frac{2}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{20}+0+\frac{-1}{5}=\frac{7-4}{20}=\frac{3}{20}\)

b ) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{-21}{23}+\frac{-125}{143}\)

\(=0+\frac{-125}{143}=-\frac{125}{143}\)

18 tháng 2 2017

bài 2

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
=\(1-\frac{1}{2004}=\frac{2003}{2004}\)