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`2/[1xx4]+2/[4xx7]+...+2/[97xx100]`
`=2/3xx(3/[1xx4]+3/[4xx7]+...+3/[97xx100])`
`=2/3xx(1-1/4+1/4-1/7+...+1/97-1/100)`
`=2/3xx(1-1/100)=2/3xx99/100=33/50`
\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)
\(=\dfrac{2}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{2}{3}.\dfrac{99}{100}\)
\(=\dfrac{33}{50}\)
\(A=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{97\times100}\right)\)
\(A=3\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3\times\left(1-\frac{1}{100}\right)\)
\(A=3\times\frac{99}{100}\)
\(A=\frac{297}{100}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+......+\frac{3^2}{97.100}\)
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Đặt \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
Ta có: \(S=\frac{3}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{97.100}\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)
\(S=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=3.S=3.\frac{99}{100}=\frac{297}{100}\)
1.
=0+0-0-0+0+0-0-0+0
=0
2.
\(=2\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{94.97}+\dfrac{1}{97.100}\right)\)
\(=2.\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{2}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}.\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}.\dfrac{99}{100}=\dfrac{2.99}{3.100}=\dfrac{1.33}{1.50}=\dfrac{33}{50}\)
bởi vì các số nào nhân với 0 cũng bằng 0 em ạ
hoặc dùng cách sau :
=0.(1+2-3-4+5+6-7-8+9)
=0.1
=0
a/ \(\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=3\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\right)\)
\(=3\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(=3\left(\dfrac{1}{11}-\dfrac{1}{15}\right)\)
\(=\dfrac{4}{55}\)
b/ \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(=\dfrac{2}{3}\)
c/ \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+.....+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
d/ \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.....+\dfrac{3}{100.103}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{1}{2}-\dfrac{1}{103}\)
\(=\dfrac{101}{206}\)
e/ Đặt :
\(A=\dfrac{1}{1.5}+\dfrac{1}{5.10}+....+\dfrac{1}{95.100}\)
\(\Leftrightarrow5A=\dfrac{5}{1.5}+\dfrac{5}{5.10}+....+\dfrac{5}{95.100}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+....+\dfrac{1}{95}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{100}:5=\dfrac{99}{500}\)
Dấu . là dấu nhân nhé <3
1. 0,1 + 0,2 - 0,3 - 0,4 + 0,5 + 0,6 - 0,7 - 0,8 + 0,9
= ( 0,1 + 0,9 ) + ( 0,2 - 0,8 ) - ( 0,3 + 0,7 ) - ( 0,4 - 0,8 ) + 0,5
= 1 + ( - 0,6 ) - 1 - ( 0,2 ) + 0,5
= 1 + ( - 0,6 ) - 1 + 0,2 + 0,5
= [ 1 + ( - 0,6 ) ] - ( 1 - 0,2 - 0,5 )
= 0,4 - 0,3 = 0,1
\(S=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{94\times97}+\frac{1}{97\times100}\)
\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\times\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\times\frac{99}{100}\)
\(S=\frac{33}{100}\)
Bài 1:
$M=3.4.5+4.5.6+...+13.14.15$
$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$
$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$
$=-2.3.4.5+13.14.15.16=43560$
$M=43560:4=10890$
Bài 2:
a.
$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$
$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$
$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$
$=1-\frac{1}{100}=\frac{99}{100}$
$M=\frac{99}{100}:3=\frac{33}{100}$
\(\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\)
\(=2\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{97\cdot100}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\)