Tính bằng cách hợp lí:B=(\(\frac{1.2+2.3+...+99.100}{99.100.101}\)):\(\frac{1}{3}\)
Những câu hỏi liên quan
TÍNH TỔNG:
\(S=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
Đúng 0
Bình luận (0)
M = \(\frac{1}{1.2}\)-\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3}\)-\(\frac{1}{2.3.4}\)+...+\(\frac{1}{99.100}-\frac{1}{99.100.101}\)
Tính bằng cách hợp lý
a) \(\frac{-1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{99.100}\)
b)\(\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{96.100}\)
\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(F=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{47.48.49.50}\)
Tính
\(C=1+\frac{1}{\left(-3\right)}+\frac{1}{\left(-3\right)^2}+....+\frac{1}{\left(-3\right)^{2015}}\)
Tính E = 1/1.2-1/1.2.3+1/2.3-1/2.3.4+1/3.4-1/3.4.5+...+1/99.100-1/99.100.101
\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{5049}{20200}\)
Suy ra \(E=A-B=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
\(\frac{14949}{20200}\)
tính \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
đặt cái trên là A
=> A=\(\frac{2-1}{1.2}+\frac{3-2}{3.2}+..+\frac{100-99}{100.99}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{100}\)
=1-1/100=99/100
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\) tính
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+........+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Chúc bạn học tốt!!!!!!
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Đúng 0
Bình luận (0)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
A=1-1/100
A=99/100
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
12 tháng 3 2017 lúc 8:11
Tính : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(5x-5x\)
\(????????\)
\(\text{Bn mún hỏi dj v~~~~}\)
\(5x-5x=0\)
~~~~~~~~~~~
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính \(A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
A=\(\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Đúng 0
Bình luận (0)
\(A=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
Đúng 0
Bình luận (0)
Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(A=\frac{99}{100}.\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời