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Đặt : \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(A=\frac{100}{101}\cdot\frac{1}{2}=\frac{50}{101}\)
Ta có:
a)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
b)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{210}\)
\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+...+\(\dfrac{1}{182}\)+ 210
A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}\)+...+ \(\dfrac{1}{13\times14}\)+ 210
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+\(\dfrac{1}{13}\) - \(\dfrac{1}{14}\) + 210
A = 1 - \(\dfrac{1}{14}\) + 210
A = 211 - \(\dfrac{1}{14}\)
A = \(\dfrac{2953}{14}\)
: A = 1/6+1/12+1/20+1/30+.........+1/210
A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/14.15
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/14 - 1/15
A = 1/2 - 1/15
A = 13/30
Ta có: 1/2= 1/1- 1/2
1/6= 1/2 - 1/3
1/12= 1/3- 1/4
...
1/30= 1/5 - 1/6
1/42= 1/6 - 1/7
Thay vào tổng kia: 1/2+1/6+...+1/30+1/42= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/5 - 1/6 + 1/6 - 1/7 = 1/2 - 1/7= 5/14
Chúc bạn học tốt. Thân!
\(B=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{210}\)
\(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{420}\)
\(=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{20.21}\)
\(=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=6\left(1-\frac{1}{21}\right)\)
\(=6.\frac{20}{21}=\frac{40}{7}\)
\(25\frac{2}{13}-\left(\frac{15}{17}+15\frac{2}{3}\right)=\frac{327}{13}-\left(\frac{15}{17}+\frac{47}{3}\right)\)
\(=\frac{327}{13}-\frac{844}{51}\)
\(=\frac{5705}{663}\)
\(\frac{5}{30}+\frac{15}{90}+\frac{250}{150}+\frac{350}{210}+\frac{45}{270}=\frac{1}{6}+\frac{1}{6}+\frac{5}{3}+\frac{5}{3}+\frac{1}{6}\)
\(=\frac{3}{6}+\frac{10}{3}\)
\(=\frac{1}{2}+\frac{10}{3}\)
\(=\frac{23}{6}\)
\(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{17}.2\frac{4}{9}=\frac{34}{11}.\frac{27}{46}.\frac{23}{17}.\frac{22}{9}\)
\(=\frac{68}{9}.\frac{27}{34}\)
\(=6\)
\(\frac{7}{10}-\frac{3}{10}x\frac{1}{10}+\frac{4}{25}\)
\(=\frac{7}{10}-\frac{3}{100}+\frac{4}{25}\)
\(=\frac{70}{100}-\frac{3}{100}+\frac{4}{25}\)
\(=\frac{67}{100}+\frac{4}{25}\)
\(=\frac{67}{100}+\frac{16}{100}\)
\(=\frac{83}{100}=0,83\)
\(\frac{6}{11}x\frac{3}{7}+\frac{3}{7}x\frac{5}{11}\)
\(=\frac{18}{77}+\frac{15}{77}\)
\(=\frac{33}{77}=\frac{3}{7}\)
\(49x\left(311-4807:23\right)+98210\)
\(=49x\left(311-209\right)+98210\)
\(=49x102+98210\)
\(=4998+98210\)
\(=103208\)
a) \(\frac{7}{10}-\frac{3}{10}\times\frac{1}{10}+\frac{4}{25}\)
\(=\frac{2}{5}\times\frac{1}{10}+\frac{4}{25}\)
\(=\frac{1}{25}+\frac{4}{25}\)
\(=\frac{5}{25}=\frac{1}{5}\)
b)\(\frac{6}{11}\times\frac{3}{7}+\frac{3}{7}\times\frac{5}{11}\)
\(=\frac{3}{7}\times\left(\frac{6}{11}+\frac{5}{11}\right)\)
\(=\frac{3}{7}\times1\)
\(=\frac{3}{7}\)
c)\(49\times\left(311-4807:23\right)+98210\)
\(=49\times\left(311-209\right)+98210\)
\(=49\times102+98210\)
\(=4998+98210\)
\(=103208\)
tổng trên sẽ là:
1/1+1/3+1/210=3/321
đáp số:3/321