Bài 1:
a)-3/5.-1/2
b)-4/7.2/3
c)-333/151.-302/111
d)-147.1/7
Bài 2 :1/1.1/2+1/2.1/3+1/3.1/4+....+1/998.1/999+1/999.1/1000
Những câu hỏi liên quan
3x.(1/1.1/2+1/2.1/3+1/3.1/4+1/4.1/5+1/5.1/6)=3/4
tính nhanh 1/1.1/2+1/2.1/3+1/3.1/4+........+1/9991000
1/1.2+1/2.3+1/3.4+1/4.5+.................+1/9990999.9991000
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.................+1/9990999-1/9991000
=1-1/9991000
=9990999/9991000
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tính nhanh:1/1.1/2+1/2.1/3+1/3.1/4+....+1/9991000
Bài 1:tính tích
a, A=(1-1/2).(1-1/3).(1-1/4)...(1.1/999).(1-1/1000)
b, B= 3/4.8/9.1/16...2499/2500
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}=\frac{1.2.3...999}{2.3.4...1000}=\frac{1}{1000}\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}=\frac{3.8.15...2499}{4.9.16....2500}=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4...50.50}=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
\(\frac{1.51}{50.2}=\frac{51}{100}\)
a. \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\)
\(A=\frac{1\cdot2\cdot3\cdot....\cdot998}{2\cdot3\cdot4\cdot....\cdot999}=\frac{1}{999}\)
Vậy \(A=\frac{1}{999}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{1000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)
\(=\frac{1.2.3.4....999}{2.3.4....1000}\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}\)
\(=\frac{3.8.15....2499}{4.9.16....2500}\)
\(=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4....50.50}\)
\(=\frac{\left(1.2.3.4.5...49\right)\left(3.4.5....51\right)}{\left(2.3.4....50\right).\left(2.3.4...50\right)}\)
\(=\frac{1.51}{50.2}=\frac{51}{100}\)
\(=\frac{1}{1000}\)
Bài 1: Tính bằng cách thuận tiện nhất.
1/1.1/2+1/2.1/3+1/3.1/4+1/4.1/5
Các bạn giúp mình nha.Cảm ơn các bạn nhiều💝
1/1.2+1/2.3+1/3.4+1/4.5+.................+1/9990999.9991000
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.................+1/9990999-1/9991000
=1-1/9991000
=9990999/9991000
thu gọn các đa thức sau:
a,2a^3.(-1/2ab).a^2b
b,-2/1/3a^3c^2.1/7ac^2.6abc
c,2ab.4/3a^2b^4.7abc
d,2y.3y^2.d^2y^2
e,(-2/1/3.cd).(1/1/4c^2d).(-5/6cd)^2
g,(1/2a.1/4a^2.1/8^3)^2.2b.4b^2-8b^3
tính hợp A=1/2+1/3-1/4:(1/2+1/3-1/4)-1/2.1/3.1/4
Bài 1:D=1000+998+996+...+2-999-997-995-1
E=100+98+96+94+...+2-97-95-93-...-1
G=1-3+5-7+...+2005-2007+2009
H=1-3+5-7+...+2001-2003+2005
I=1+0-3-4+5+6-7-8+9+...-299+300+301+302
K=1-2+3-4+...+2017-2018+2019
L=1-2-3+4+5-6-7+...+1997+1998-1999+2000+2001
M=1-2+3-4+5-6+...+99-100+101
Bài 2:A=1+2+3+...+x
B=2+4+6+...+2.x
C=1+3+5+7+...+(2x-1)
1/.1/1.1/2+1/2.1/3+1/3.1/4+1/4.1/5
2/.1/2+1/6+1/12+...+1/10100
3/.A = 2/1.3+2/3.5+2/5.7+...+2/99.101
4/.A = 1/1.3+1/3.5+1/5.7+...+1/99.101
tính bằng cách thuận tiện nhất ( làm nhanh trước 5h nha , nếu ai làm được thì cho 100 tick , thật đó và trình bày cách diễn giải nha
3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
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4)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{100}{101}\)
A = \(\frac{50}{101}\)
2, đặt tên biểu thức trên là A. Ta có :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
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