\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\).

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)

\(2A=\frac{1}{1}-\frac{1}{201}\)

\(2A=\frac{201-1}{201}\)

\(2A=\frac{200}{201}\)

\(A=\frac{200}{201}:2\)

\(A=\frac{200}{402}\)

Đáp số là 100/201

hông biết đúng không nữa?

G=7/2.(2/1.3+2/3.5+2/5.7+...+2/199.201)

G=7/2.(1-1/3+1/3-1/5+1/5-1/7+.....+1/199.201

G=7/2.(1-1/201)

G=7/2.200/201

G=1400/402

G=700/201

\(G=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{199.201}\)

\(G=7\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.200}\right)\)

\(G=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)

\(G=\frac{7}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)

\(G=\frac{7}{2}\left(1-\frac{1}{201}\right)\)

\(G=\frac{7}{2}.\frac{200}{201}\)

\(G=\frac{700}{201}\)

= 700/201

\(D=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{199.201}\)

\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)

\(D=\frac{3}{2}.\frac{200}{201}\)

\(D=\frac{100}{67}\)

#)Giải :

\(D=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{199.201}\)

\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)

\(D=\frac{3}{2}\times\frac{200}{201}\)

\(D=\frac{100}{67}\)

\(\frac{2}{3}D=\frac{2}{1\times3}+\cdot\cdot\cdot+\frac{2}{199\times201}\)

\(\frac{2}{3}D=1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{199}-\frac{1}{201}\)

\(\frac{2}{3}D=1-\frac{1}{201}\)

\(\frac{2}{3}D=\frac{200}{201}\)

\(D=\frac{200}{201}:\frac{2}{3}=\frac{200\times3}{201\times2}=\frac{300}{201}\)

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

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

2/1.3 + 2/3.5 + 2/5.7 +...+ 2/97.99 
=(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/97-1/99) 
=1-1/99=98/99 

 

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=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100

A=1-1/100

A=99/100

\(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)\(=\left(1-\frac{1}{101}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)=\left(\frac{101}{101}-\frac{1}{101}\right)+0+0+...+0=\frac{100}{101}\)Chúc bạn học tốt!^_^