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Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
=> 2A = \(2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
= \(2\times\frac{1}{2}+2\times\frac{1}{2^2}+2\times\frac{1}{2^3}+...+2\times\frac{1}{2^{50}}\)
= \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
A = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{50}}\)
= \(1-\frac{1}{2^{50}}\)
Vậy \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)= \(1-\frac{1}{2^{50}}\)
a, \(\dfrac{4}{7}\). \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\)
\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\):\(\dfrac{4}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\)
b, \(\dfrac{a}{b}\) + \(\dfrac{2}{3}\).\(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) + \(\dfrac{2}{9}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\) - \(\dfrac{2}{9}\)
\(\dfrac{a}{b}\) = \(\dfrac{4}{9}\)
c, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}.\)\(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{13}{21}\)
d, \(\dfrac{11}{13}\): \(\dfrac{a}{b}\): \(\dfrac{2}{3}\) = 2\(\dfrac{7}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\):\(\dfrac{2}{3}\) = \(\dfrac{33}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{33}{13}\) \(\times\) \(\dfrac{2}{3}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{11}{13}\) : \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{1}{2}\)
\(A=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)
\(A=\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)
\(A=\frac{59}{15}-\left(\frac{21}{2}-\frac{14}{3}\right):\frac{7}{4}\)
\(A=\frac{59}{15}-\frac{35}{6}:\frac{7}{4}\)
\(A=\frac{59}{15}-\frac{10}{3}\)
\(A=\frac{3}{5}\)
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
1/2.2 + 1/3.3 + 1/4.4 +....+ 1/99.99 + 1/100.100
= 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/98.99 + 1/99.100
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100
= 1/1 - 1/100
= 99/100
a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)
A = 2 + 22 + 23 + 24 + ... + 216
2A = 22 + 23 + 24 + 25 + ... + 217
2A - A = ( 22 + 23 + 24 + 25 + ... + 217) - ( 2 + 22 + 23 + 24 + ... + 216)
A = 217 - 2
b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56
1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8
1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
1/2 x B = 1 - 1/8 = 7/8
B = 7/8 : 1/2
B = 7/8 x 2 = 7/4
a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)
A = 2 + 22 + 23 + 24 + ... + 216
2A = 22 + 23 + 24 + 25 + ... + 217
2A - A = ( 22 + 23 + 24 + 25 + ... + 217) - ( 2 + 22 + 23 + 24 + ... + 216)
A = 217 - 2
b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56
1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8
1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
1/2 x B = 1 - 1/8 = 7/8
B = 7/8 : 1/2
B = 7/8 x 2 = 7/4