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Ta có\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{10100}\right)\)

\(A=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)

\(A=\left(1+1+1+...+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)

           100 số 1

\(A=100+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=100+\left(1-\frac{1}{101}\right)\)

\(A=100+1-\frac{1}{101}\)

\(A=101-\frac{1}{101}< 101=B\)

\(\Rightarrow A< B\)

Vậy A<B

Học tôt nha

7 tháng 9 2019

Câu hỏi của Lê Tiến Cường - Toán lớp 6 - Học toán với OnlineMath

16 tháng 8 2023

\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+...+\dfrac{2}{10100}\)

\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{100\times101}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)

\(A=2\times\dfrac{99}{202}\)

\(A=\dfrac{99}{101}\)

28 tháng 6 2015

    1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100

= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101

= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101

= 1/1 - 1/101

= 100 /101

28 tháng 6 2015

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{9900}+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

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

=\(\frac{100}{101}\)

28 tháng 6 2015

Mik trả lời ở bài dưới rồi đó.

28 tháng 6 2015
1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100 = 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101 = 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101 = 1/1 - 1/101 = 100 /101
10 tháng 8 2022

 

1.3.77−1​+3.7.99−3​+7.9.1313−7​+9.13.1515−9​+\frac{19-13}{13.15.19}+13.15.1919−13​

=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31​−3.71​+3.71​−7.91​+7.91​−9.131​+9.131​−13.151​+13.151​−15.191​

=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31​−15.191​=28595​−2851​=28594​

b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61​.(1.3.76​+3.7.96​+7.9.136​+9.13.156​+13.15.196​)

làm giống như trên

c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81​.(1.2.31​+2.3.41​+3.4.51​+...+48.49.501​)

=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161​.(1.2.32​+2.3.42​+3.4.52​+...+48.49.502​)

=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161​.(1.2.33−1​+2.3.44−2​+3.4.55−3​+...+48.49.5050−48​)

=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161​.(1.21​−2.31​+2.31​−3.41​+3.41​−4.51​+...+48.491​−49.501​)

=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161​.(21​−24501​)=161​.(24501225​−24501​)=4900153​

d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75​.(1.5.87​+5.8.127​+8.12.157​+...+33.36.407​)

=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75​.(1.5.88−1​+5.8.1212−5​+8.12.1515−8​+...+33.36.4040−33​)

=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75​.(1.51​−5.81​+5.81​−8.121​+8.121​−12.151​+...+33.361​−36.401​)

=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75​.(51​−14401​)=75​.(1440288​−14401​)=28841​

P/S: . là nhân nha

29 tháng 6 2015

a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{100}-\frac{1}{101}\)

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

=\(\frac{100}{101}\)

b,\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)

=\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{256}-\frac{1}{512}\right)\)

=\(1-\frac{1}{512}\)

=\(\frac{511}{512}\)

29 tháng 6 2015

\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}+\frac{1}{10100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

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