Tính nhanh (1+1/2)*(1+1/3)*(1+1/4)...*(1+1/9)*(1+1/10)
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VM
2
LS
10 tháng 7 2018
Đặt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)
\(A=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{10.11}{2}}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(A=2\cdot\frac{9}{22}=\frac{9}{11}\)
Vậy A = \(\frac{9}{11}\)
ZC
1
29 tháng 4 2016
A = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\)+...+\(\frac{1}{9.10}\)
= (\(\frac{1}{6}\) + \(\frac{1}{12}\)) + ( \(\frac{1}{20}\)+\(\frac{1}{30}\) ) + ( \(\frac{1}{42}\)+\(\frac{1}{56}\) ) + ( \(\frac{1}{72}\)+ \(\frac{1}{90}\) )
= \(\frac{1}{4}\) + \(\frac{1}{12}\) + \(\frac{1}{24}\) + \(\frac{1}{40}\)
= (\(\frac{1}{4}\)+ \(\frac{1}{12}\) ) + ( \(\frac{1}{24}\)+ \(\frac{1}{40}\) )
= \(\frac{1}{3}\) + \(\frac{1}{15}\)
= \(\frac{2}{5}\)
18 tháng 3 2017
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)......\left(1+\frac{1}{9}\right)\left(1+\frac{1}{10}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{10}{9}.\frac{11}{10}=\frac{3.4.5......10.11}{2.3.4.....9.10}\)
\(=\frac{11}{2}\)
15 tháng 3 2017
bạn ơi tại sao bạn lại ra kết quả nh vậyke chi tiết hơn được không vậy
9 tháng 8 2023
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)