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-1 + 3- 5 + 7-...-97 + 99 ( có 50 số hạng )
= (3 - 1)+(7 - 5)+ ... +(99 - 97) ( có 25 nhóm )
= 2 + 2 + ... + 2 (có 25 số hạng 2)
= 25.2
=50
Đặt BT trên là A
\(\frac{2}{5}.A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(\frac{2}{5}.A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}.\frac{5}{2}=\frac{250}{101}\)
Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{5}{2}.\frac{100}{101}=\frac{5.50}{101}=\frac{550}{101}\)
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
1. Tính hợp lí
a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)
\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)
\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)
\(=1+\dfrac{-7}{19}\)
\(=\dfrac{12}{19}\)
b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)
\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)
\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)
\(=\dfrac{3}{5}.-1\)
\(=\dfrac{-3}{5}\)
d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)
\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)
\(=\dfrac{7}{4}.1\)
\(=\dfrac{7}{4}\)
Chúc bạn học tốt
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{100}\right).\)
\(=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(G=5+5^3+5^5+5^7+...+5^{99}\)
=> \(25G=5^3+5^5+5^7+5^9+...+5^{101}\)
=> \(24G=5^{101}-5=>G=\frac{5^{101}-5}{24}\)
G = 5 + 5^3 + 5^5 + ...+ 5^99
=> 52G = 5^3 + 5^5 + 5^7 + ...+ 5^101
=> 52G-G = 5^101 - 5
24G = 5^101 - 5
\(G=\frac{5^{101}-5}{24}\)