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1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000
= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000
= 1/10 - 1/1000
= 100/1000 - 1/1000
= 99/1000
1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000
= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000
= 1/10 - 1/1000
= 100/1000 - 1/1000
= 99/1000
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
= 7/10-7/11+7/11-7/12+7/12-7/13+...+7/69-7/70
=7/10-7/70
=42/70
k mk nha
=[2013*2014*+2014*2015+2015*2016]*[\(1\frac{1}{3}-1\frac{1}{3}\)]
=A*0
=0
nhớ k nha mấy friends/
\(1\cdot\frac{1}{15}\cdot1\frac{1}{16}\cdot1\frac{1}{17}\cdot....\cdot1\frac{1}{2016}\cdot1\frac{1}{2017}\)
\(=\frac{1}{15}\cdot\frac{17}{16}\cdot\frac{18}{17}\cdot....\cdot\frac{2017}{2016}\cdot\frac{2018}{2017}\)
\(=\frac{1}{15}\cdot\frac{1}{16}\cdot2018\)
Dấu "." là dấu nhân nhé bn! phần còn lại bn làm tiếp nha
=(3/2)x(4/3)x....x(100/99)
=(3x4x5x...x100)/(2x3x...x99)
=100/2
=50
a) \(\frac{3}{5}+25-\frac{1}{5}=\left(\frac{3}{5}-\frac{1}{5}\right)+25=\frac{2}{5}+25=\frac{2}{5}+\frac{125}{5}=\frac{127}{5}\)
b) \(13\times3\times32,27+67,63\times39=39\times32,27+67,63\times39\)
\(=39\times\left(32,27+67,63\right)\)
\(=39\times99,9=3196,8\)
(Bạn xem lại đề nhé)
c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times....\times\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{99}{100}\)
\(=\frac{1\times2\times3\times...\times99}{2\times3\times4\times....\times100}=\frac{1}{100}\)
a, \(\frac{3}{5}+25-\frac{1}{5}=\frac{127}{5}\)
b, \(\text{13 x 3 x 32,27 + 67,63 x 39 =}3896,1\)
............................
thank for watching me do homework
có trong câu hỏi tương tự rùi mà bn
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
=\(\frac{1.2.3.4}{2.3.4.5}\)
=\(\frac{1}{5}\)
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)\)
= \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}\)
= \(\frac{1x2x3x4}{2x3x4x5}=\frac{1}{5}\)
mới đầu mình cũng định làm như lê thành đạt , nhưng x là dấu nhân thì đâu thể làm theo cách đấy .
vd:1/(10x11)x1/(11x12) khác vs 1/(10x11)+1/(11x12)
hay cậu cho đề sai ????