\(D=\frac{3}{3x4}+\frac{3}{4x5}+......+\frac{3}{98x99}+\frac{3}{99x100}\)
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TN
11 tháng 3 2016
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)
\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
NP
2
14 tháng 5 2019
D= 3/3x4+3/4x5+...+3/99x100
D=3x(1/3x4+1/4x5+....+1/99x100)
D=3x(1/3-1/4+1/4-1/5+...+1/99-1/100)
D=3x(1/3-1/100)
D=3x(100/300-3/300)
D=3x97/300=97/100
Nhớ tk cho mình nha
14 tháng 5 2019
\(D=\frac{3}{3\times4}+\frac{3}{4\times5}+...+\frac{3}{98\times99}+\frac{3}{99\times100}\)
\(=3\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)
\(=3\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3\times\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(=3\times\frac{97}{300}\)
\(=\frac{97}{100}\)
15 tháng 6 2016
1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
VP
1
HT
2 tháng 8 2016
\(\frac{3}{1.2}+\frac{3}{2.3}+........+\frac{3}{99.100}\)
\(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=\frac{3.99}{100}=\frac{297}{100}\)
21 tháng 9 2020
2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{1.50}{100}-\frac{1}{100}=\frac{50-1}{100}=\frac{49}{100}\)
HM
12 tháng 8 2017
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
TM
1 tháng 8 2017
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
27 tháng 4 2018
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1/1-1/100
=100/100-1/100
=99/100
27 tháng 4 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)
= \(\frac{99}{100}\)
~~~
#Sunrise
\(D=\frac{3}{3x4}+\frac{3}{4x5}+.....+\frac{3}{99x100}.\)
\(D=3x\left(\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}+\frac{1}{99x100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(D=1-\frac{3}{100}\)
\(D=\frac{97}{100}\)
\(D=\frac{3}{3x4}+\frac{3}{4x5}+.........+\frac{3}{98x99}+\frac{3}{99x100}\)
\(D=3x\left(\frac{1}{3x4}+\frac{1}{4x5}+...........+\frac{1}{98x99}+\frac{1}{99x100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\right)\)
\(D=3x\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(D=\frac{3x97}{100}\)
\(D=\frac{291}{100}\)