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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
LL
0
DT
1
K
27 tháng 6 2015
\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)
\(\frac{99}{20}-2x=\frac{49}{99}\)
\(2x=\frac{99}{20}-\frac{49}{99}\)
\(2x=\frac{8821}{1980}\)
\(x=\frac{8821}{1980}:2\)
\(x=\frac{8821}{3960}\)
H
15 tháng 5 2017
Có:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(1-0-0-0...-0-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{5}{11}:\dfrac{1}{2}=\dfrac{10}{11}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{10}{11}\)
\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{11}\)
\(\Rightarrow x+2=11\)
\(\Rightarrow x=11-2=9\)
Vậy x = 9.
Chúc bạn học tốt!
15 tháng 5 2017
1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2)
= 1/2.(1/1 - 1/3) + 1/2.(1/3 - 1/5) + 1/2.(1/5 - 1/7) + ... + 1/2.(1/x -1/x+2)
= 1/2.(1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 )
= 1/2.(1/1 - 0 - 1/x+2 )
= 1/2 . ( 1/1 - 1/x+2 )
= 1/2 . ( x+2/x+2 - 1/x+2 )
= 1/2 . x+1/x+2
Mà 1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2) = 5/11
=> 1/2 . x+1/x+2 = 5/11
=> x+1/x+2 = 5/11 : 1/2
=> x+1/x+2 = 10/11
=> x+1/x+2-1 = 10/11-1
=> x+1/x+1 = 10/10
=> x + 1 = 10
=> x = 10 - 1
=> x = 9
Vậy x = 9
MC
4 tháng 5 2016
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
L
0
TV
1
OY
5 tháng 12 2021
Đặt biểu thức trên là A
\(A=5+55+555+...+555...5\) (2011 c/s 5)
\(\dfrac{5}{9}A=9+99+999+...+9999...9\) (2011 c/s 9)
\(\dfrac{5}{9}A=\left(10-1\right)+\left(100-1\right)+...+\left(100...0-1\right)\) (2011 c/s 0)
\(\dfrac{5}{9}A=10+100+...+100...0-\left(1+1+...+1\right)\)
\(\dfrac{5}{9}A=10+10^2+...+10^{2011}-2011\)
Đến đây em tự giải nốt
a/
\(A=5\left(1+11+111+...+111...1\right)\) (1999 chữ số 1)
\(A=5\left(\dfrac{10-1}{9}+\dfrac{100-1}{9}+\dfrac{1000-1}{9}+...+\dfrac{1000...0-1}{9}\right)\) (1999 chữ số 0)
\(A=5\left(\dfrac{10+10^2+10^3+...+10^{1999}-1999}{9}\right)\)
Đặt
\(B=10+10^2+10^3+...+10^{1999}\)
\(10B=10^2+10^3+10^4+...+10^{2000}\)
\(9B=10B-B=10^{2000}-10\)
\(B=\dfrac{10^{2000}-10}{9}=\dfrac{10\left(10^{1999}-1\right)}{9}=\dfrac{10.999...9}{9}=10.111...1\) (1999 chữ số 1)
\(\Rightarrow A=5\left(\dfrac{10.111...1-1999}{9}\right)\) (1999 chữ số 1)
b/
\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{17.19}\)
\(2C=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{19-17}{17.19}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{17}-\dfrac{1}{19}=\)
\(=1-\dfrac{1}{19}=\dfrac{18}{19}\Rightarrow C=\dfrac{18}{19}:2=\dfrac{9}{19}\)