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
NT
9
28 tháng 4 2015
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
HD
1
TT
14 tháng 4 2016
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
3 tháng 5 2015
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2
=(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2
=(1/2-1/2010).2
=1004/1005
MH
11 tháng 2 2022
\(A=1\times3+3\times5+5\times7+...+99\times101\)
\(=1\left(1+2\right)+3\left(3+2\right)+5\left(5+2\right)+...+99\left(99+2\right)\)
\(=\left(1^2+3^2+5^2+...+99^2\right)+2\left(1+3+5+...+99\right)\)
Ta có:
\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)
⇒ \(A=\left(1^2+2^2+3^2+...+100^2\right)-2^2\left(1^2+2^2+3^2+...+50^2\right)+2\left(1+3+5+...+99\right)\)
\(=\dfrac{100.101.201}{6}+\dfrac{4.50.51.101}{6}+\dfrac{\left(99+1\right).\left[\left(99-1\right):2+1\right]}{2}\)
\(=338350-171700+5000\)
\(=166650+5000=171650\)
11 tháng 2 2018
Bài 1:
Ta có:
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b, Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
\(\Rightarrow\frac{2}{5}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
Từ (a) \(\Rightarrow\frac{2}{5}A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:\frac{2}{5}=\frac{100}{101}.\text{5/2}=\frac{250}{101}\)
Bài 2:
Đặt \(\left(2n+1;3n+2\right)=d\left(d\inℕ^∗\right)\)
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}}\)
\(\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)\Rightarrow d=1\)
\(\Rightarrow\left(2n+1;3n+2\right)=1\)
\(\Rightarrow\frac{2n+1}{3n+2}\)là phân số tối giản
NJ
11 tháng 2 2018
1. Giải
a, \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=2.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\right)\)
\(=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=5.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{5.100}{2.101}=\frac{500}{202}=\frac{250}{101}\)
2. Giải
Gọi ước chung lớn nhất của 2n + 1 và 3n + 2 là d (d thuộc N*)
=> 2n + 1 \(⋮\)d ; 3n + 2 \(⋮\)d
=> 3(2n + 1) \(⋮\)d ; 2(3n + 2) \(⋮\)d
=> 6n + 3 \(⋮\)d , 6n + 4 \(⋮\)d
=> (6n + 4) - (6n + 3) \(⋮\)d
=> 1 \(⋮\)d
=> d = 1
Vậy \(\frac{2n+1}{3n+2}\)là phân số tối giản
Đặt A = 1.3+3.5+5.7+ ...+ 99.101
ta có : 6A = 1.3.6+3.5.6+5.7.6 +...+99.101.6
= 1.3.(5+1) + 3.5.( 7-1) +5.7.(9-3) +...+99.101.(103-97)
=1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+99.101.103-97.99.101
=1.3+99.101.103
A =1029900 :6 = 171650
Answer:
Ta đặt \(A=1.3+3.5+5.7+...+99.101\)
\(\Rightarrow6A=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)\)
\(\Rightarrow6A=1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+99.101.103-97.99.101\)
\(\Rightarrow6A=3+99.101.103\)
\(\Rightarrow6A=1029900\)
\(\Rightarrow A=1029900:6=171650\)