Tính:
a) A=1.2+3.4+5.6+...+21.22
b) B=1.3+5.7+9.11+...+41.43
c) C=12.2+22.3+32.4+...+212.22
K
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SK
2
LN
26 tháng 4 2017
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
=1\(-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(\dfrac{47}{60}\)
B=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)=
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}+\dfrac{1}{101}\)
=\(1-\dfrac{1}{101}\)
=\(\dfrac{100}{101}\)
25 tháng 4 2017
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
= \(\dfrac{47}{60}\)
B= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= 2\(\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{200}{101}\)
31 tháng 1 2021
Bài 5:
a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)
\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)
\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)
hay A=330
Vậy: A=330
T
9 tháng 9 2017
a)Đặt \(A=\dfrac{6}{1.4}+\dfrac{6}{4.7}+\dfrac{6}{7.10}+...+\dfrac{6}{97.100}\)
\(3a=3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+...+\dfrac{3}{97}-\dfrac{3}{100}\)
\(=3-\dfrac{3}{100}\)
\(=\dfrac{297}{100}\)
b)Đặt \(B=\dfrac{4}{1.3}+\dfrac{16}{3.5}+\dfrac{36}{5.7}+...+\dfrac{9604}{97.99}\)
\(=2b=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(2b=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{97}-\dfrac{2}{99}\)
\(2b=2-\dfrac{2}{99}=\dfrac{198}{99}-\dfrac{2}{99}=\dfrac{196}{99}\)
c) Tương tự! Bạn tự làm nhé!
10 tháng 12 2016
Bài 1 : Ta có : a = 1.2 + 2.3 + 3.4 + ....... + 99.100
=> 3a = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...... + 99.100.(101 - 98)
=> 3a = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101
=> 3a = 99.100.101
=> a = \(\frac{99.100.101}{3}=333300\)
CB
1
NT
2
12 tháng 8 2018
B = 1.2+2.3+3.4+...+99.100
B=1.100
B=100
C=1.3+2.4+3.5+4.6+...+9.11
C=1.(2+1)+2.(3+1)+3.(4+1)+4.(5+1)+...+9.(10+1)
C=1.2+1+2.3+1+3.4+1+4.5+1+...+9.10+1
C=(1.2+2.3+3.3+4.5+...+9.10)+(1+1+1+1+..+1)
C=1.10+10
C=10+10
C=20
12 tháng 8 2018
a) B = 1.2+2.3+3.4+..+99.100
=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3
3B = 1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5-2.3.4+...+99.100.101-98.99.100
3B = (1.2.3+2.3.4+3.4.5+..+99.100.101) - (1.2.3+2.3.4+...+98.99.100)
3B = 99.100.101
\(B=\frac{99.100.101}{3}=333300\)
b) C = 1.3+2.4+3.5+4.6+...+9.11
C = (2-1).(2+1)+(3-1).(3+1) + (4-1).(4+1)+(5-1).(5+1)+...+(10-1).(10+1)
C = 22 - 1 + 32 - 1 + 42 - 1 + 52 - 1 +...+102 - 1
C = (22+32+42+52+...+102) -(1+1+...+1)
...
NM
1
28 tháng 9 2015
A = 1.2. + 2.3 + 3.4 + ... + 99.100
3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 2.3.1 + ... + 99.100.101 - 99.100.98
3A = 99.100.101
3A = 999900
A = 333300