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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
S = \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+.......+\frac{5}{17.19}\)
S : 5 = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{17.19}\)
S : 5 = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}+.......+\frac{1}{17}-\frac{1}{19}\)
=> S : 5 = 1 - \(\frac{1}{19}=\frac{19}{19}-\frac{1}{19}=\frac{18}{19}\)
=> S = \(\frac{18}{19}x5=\frac{90}{19}\)
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) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
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
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{21.23}\)
\(=5-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+\frac{5}{5}-\frac{5}{7}+...+\frac{5}{21}-\frac{5}{23}\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=5\left(1-\frac{1}{23}\right)\)
\(=5.\frac{22}{23}\)
\(=\frac{110}{23}\)
\(A=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{21}-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\frac{22}{23}\)
\(A=\frac{55}{23}\)
\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)
\(\frac{99}{20}-2x=\frac{49}{99}\)
\(2x=\frac{99}{20}-\frac{49}{99}\)
\(2x=\frac{8821}{1980}\)
\(x=\frac{8821}{1980}:2\)
\(x=\frac{8821}{3960}\)
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
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
\(A=\dfrac{5}{3.5}+\dfrac{5}{5.7}+\dfrac{5}{7.9}+...+\dfrac{5}{97.99}\)
\(A=\dfrac{5}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
Nhận xét:
\(\dfrac{2}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};\dfrac{2}{5.7}=\dfrac{1}{5}-\dfrac{1}{7};...;\dfrac{2}{97.99}=\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Rightarrow A=\dfrac{5}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{5}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{5}{2}.\dfrac{32}{99}=\dfrac{80}{99}\)
Vậy A=\(\dfrac{80}{99}\)
A= 5/2. (2/3.5+2/5.7+2/7.9+...+/97.99) A= 5/2. ( 1/3-1/5+1/5-1/7+1/7-1/9+...+/97+1/99) A= 5/2. ( 1/3-1/99) A= 5/2.32/99 A=80/99 B=5/2. (2/1.3+2/3.5+2/5.7+...+2/59.61) B= 5/2. (1-1/3+1/3-1/5+1/5-1/7+...+1/59-1/61) B= 5/2. (1- 1/61) B = 5/2. 60/61 B = 300/122=150/61