NH
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
3 tháng 1 2018
A = (13x+5a)+(21b-3b) = 18a+18b = 18.(a+b) = 18.100 = 1800
B = (1+100).100 : 2 = 5050
Tk mk nha
3 tháng 1 2018
A=13a+21b+5a-3b
A=(13a+5a)+(21b-3b)
A=18a+18b
A=18.(a+b)
tha a+b+100ta được:
A=18.100
A=1800
B=1+2+3+...+99+100
số số hạng của tổng Blà(100-1):1+1=100
vậy B=(100+1).100:2=5050
C=1.2+2.3+3.4+...+99.100
3C=1.2.3+2.3.3+3.4.3+...+99.100.3
3C=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3C=(1.2.3+2.3.4+3.4.5+...+99.100.101)-(0.1.2+1.2.3+2.3.4+...+98.99.100)
3C=99.100.101-0.1.2
3C=999900-0
3C=999900
C=999900:3
C=333300
13 tháng 6 2017
\(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3A=99\cdot100\cdot101\Rightarrow A=\dfrac{99\cdot100\cdot101}{3}=333300\)
\(B=1^2+2^2+3^2+...+99^2+100^2\)
\(=\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=\dfrac{2030100}{6}=338350\)
\(C=1\cdot2\cdot3+2\cdot3\cdot4+...+8\cdot9\cdot10\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+8\cdot9\cdot10\cdot\left(11-7\right)\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)
\(4C=8\cdot9\cdot10\cdot11\Rightarrow C=\dfrac{8\cdot9\cdot10\cdot11}{4}=1980\)
21 tháng 7 2017
c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4
==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)
==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11
==> 4C= 8.9.10.11=7920
==> C= 7920 :4=1980
a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3
3A=1.2.(3-0) + 2.3.(4-1)+ 3.4.(5-2)+ ... + 99.100.( 101-98)
3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)
3A= 99.100.101 - 0.1.2
3A= 999900 - 0
3A= 999900
==> A= 999900 : 3
==> A= 333300
TL
0
K
23 tháng 6 2015
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
K
23 tháng 6 2015
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
3 tháng 11 2016
C=1.2+2.3+...+99.100
3C=1.2.3+2.3.3+...+99.100.3
3C=1.2(3-0)+2.3(4-1)+...+99.100(101-98)
C=99.100.101 phần 3
C=333 300
LK
2
11 tháng 4 2017
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
11 tháng 4 2017
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
20 tháng 11 2017
a)đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100
A=1.2+2.3+3.4+4.5+...+99.100
=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101
=99.100.101=999900
=>A=999900:3=333300
Vậy A=333300
b)D=2 +4 +6 +333 300
D=4+16+36+333 300
D=20+36+333 300
D=56+333 300
D=333 356
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101
= 99.100.101
A=333300
B= (1.2 + 2.3 + 3.4 + ... + 100.101) - (1 + 2 + 3+ 4 + ... + 100)
= 333300 + 10100 - 5050
= 333300 + 5050
= 338350
A = 1*2 + 2*3 + 3*4 + ........+ 99*100
=>3A=1.2.3+2.3.3+3.4.3+...+99.100.3
<=> 3A =1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
<=> 3A =1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
<=> 3A = 99.100.101 = 999900
=> S = 333300