B=4/3.5-6/5.7+8/7.9-10/9.11+.....+2016/2015.2017-2018/2017.2019
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Những câu hỏi liên quan
\(E=\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+...+\frac{100}{99.101}\)
A=\(\dfrac{4}{3.5}-\dfrac{6}{5.7}+\dfrac{8}{7.9}-\dfrac{10}{9.11}+\dfrac{12}{11.13}-...-\dfrac{100}{99.100}\)
Tính giá trị của A
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
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B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
Tìm x biết
2.x+2/3.5+2/5.7+2/7.9+2/9.11= -2016/2017
\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
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\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{8}{33}=\frac{-2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
Số dư dài quá. Đến đây bạn tự làm tiếp nhé
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\(\text{ }=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
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4\3.5 .6\5.7......2016\2015.2017
Cm tổng trên >1\6
a) M = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/97.99
b) N = 3/5.7 + 3/7.9 + 3/9.11 + ... + 3/197.199
c) P = 1/1.2 + 2/2.4 + 3/4.7 + ... + 10/46.56
Tính:
a) M=2/3.5+2/5.7+2/7.9+...+2/97.99
b) N=3/5.7+3/7.9+3/9.11+...+3/197.199
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
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