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T
7
TN
1
20 tháng 7 2016
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{99\cdot101}\)
\(=2\cdot\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2} \cdot\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}\cdot\frac{100}{101}\)
\(=\frac{50}{101}\)
XO
26 tháng 5 2021
\(B=\frac{2^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2^2}{195.197}=2\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{195.197}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{195}-\frac{1}{197}\right)=2\left(1-\frac{1}{197}\right)=2.\frac{196}{197}=\frac{392}{197}\)
DT
4
6 tháng 3 2017
4/3.5+4/5.7+4/7.9+4/9.11
=4.(1/3.5+1/5.7+1/7.9+1/9.11)
=4.1/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
6 tháng 3 2017
4/3.5+4/5.7+4/7.9+4/9.11
=4.2/2.3.5+4.2/2.5.7+4.2/2.7.9+4.2/2.9.11
=4/2.2/3.5+4/2.2/5.7+4/2.2/7.9+4/2.2/9.11
=4/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=4/2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
NT
0
25 tháng 6 2016
a :
[( 10 - x ) . 2 + 5] : 3 - 2 = 3
( 10 - x ) x 2 + 5 = ( 3+2) x 3
( 10 - x ) x 2 + 5 = 15
10 -x = ( 15 -5 ) : 2
10 - x = 5
x = 5
CW
25 tháng 6 2016
a) \(\left[\left(10-x\right)2+5\right]:3-2=3\)
\(\left(20-2x+5\right):3-2=3\)
\(\left(20-2x+5\right):3=5\)
\(20-2x+5=15\)
\(20-2x=10\)
\(2x=10\)
\(x=5\)
b) \(6x-302=2^3\cdot5\)
\(6x=8\cdot5+302\)
\(6x=342\)
\(x=57\)
c) \(12\left(x-1\right):3=4^3-2^3\)
\(12\left(x-1\right):3=56\)
\(12\left(x-1\right)=168\)
\(x-1=12\)
\(x=13\)
NH
2
M
12 tháng 12 2019
a) \(100-\left[90+\left(50-2^3.5\right)\right]\)
\(=100-\left[90+\left(50-8.5\right)\right]\)
\(=100-\left[90+\left(50-40\right)\right]\)
\(=100-\left[90+10\right]\)
\(=100-100\)
\(=0\)
b) \(-10-\left(-72+8\right)\)
\(=-10-\left(-64\right)\)
\(=54\)
Hc tốt
12 tháng 12 2019
a) 100 - [90 + (50 - 23 . 5)]
= 100 - [90 + (50 - 8 . 5)]
= 100 - [90 + (50 - 40)]
= 100 - (90 + 10)
= 100 - 100 = 0
b) -10 - (-72 + 8)
= -10 - (-64)
= -10 + 64
= 54
#Học tốt!!!
~NTTH~
TT
1
30 tháng 9 2015
6F=1.3(5-1)+3.5(7-1)+5.7(9-3)+...99.101(103-97)
6F=1.3+1.3.5-1.3.5+3.5.7-3.5.7+.....-97.99.101+99.101.103
6F=3+99.101.103
6F=3+1029897
6F=1029900
F =1029900:6
F=171650
12 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
LT
Lương Thị Vân Anh
CTVHS
12 tháng 3 2023
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
\(6x-302=2^3.5\)
\(6x-302=8.5\)
\(6x-302=40\)
\(6x=40+302\)
\(6x=342\)
\(\Rightarrow x=57\)
\(6x-302=2^3.5\)
\(\Rightarrow6x-302=8.5\)
\(\Rightarrow6x-302=40\)
\(\Rightarrow6x=40+302=342\)
\(\Rightarrow x=342:6=57\)
Vậy x = 57