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9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
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a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
Ta có: \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{1}{10}-\frac{1}{9}-...-\frac{1}{2}-\frac{1}{1}\)
= \(\frac{9}{10}+\frac{1}{10}-\frac{1}{1}\)
= 1 - 1 = 0
Vậy kết quả của phép tính là 0
Ta có :
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
= 9/10 -( 1/90 + 1/72 + ... + 1/2)
= 9/10 - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)}
= 9/10 - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10)
= 9/10 - ( 1 - 1/10)
= 9/10 - 9/10
= 0
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
=9/10-(1/9*10+1/8*9+...+1/1*2)
=9/10-(1/9-1/10+...+1-1/2)
=9/10-(-1/10+1)=9/10-9/10=0
= 9/1.10 + 1/9.10 + 1/8.9 + 1/7.8 + 1/6.7 +1/5.6 + 1/4.5 +1/3.4 +1/2.3 + 1/1.2
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 9/1.10 ( viết ngược lại)
= 1-1/2 + 1/2 -1/3 + 1/3 +....-1/10
= 1 - 1/10
= 9/10
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\) \(+?\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\) \(+?\)
\(=>?=\frac{1}{10\cdot11}=\frac{1}{110}\)
Vậy \(?\) là \(\frac{1}{110}\)
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9\cdot10}+\dfrac{1}{9\cdot8}+\dfrac{1}{7\cdot8}+\dfrac{1}{7\cdot6}+\dfrac{1}{5\cdot6}-\dfrac{1}{5\cdot4}-\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot2}-\dfrac{1}{1\cdot2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\dfrac{9}{10}\)
\(=0\)