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)100/101

b)250/101

A)100/101

B)250/101

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

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 

mình làm câu 4 nha

Gọi d là ước chung của 2n+1 và 3n+2 (d thuộc N*)

=>(2n+1) : d và (3n+2) : d

=>3.(2n+1) :d và 2.(3n+2): d

=>(6n+3) :d và (6n+4) : d

=> ((6n+4) - (6n+3)) : d

=>1 :d => d=1

Vì d là ước chung của 2n+1/3n+2

mà d =1 => ƯC(2n+1/3n+2) =1

Vậy 2n+1/3n+2 là phân số tối giản

Tick mình nha bạn hiền .

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

Câu 1: B

\(A=1\times3+3\times5+5\times7+...+99\times101\)

\(=1\left(1+2\right)+3\left(3+2\right)+5\left(5+2\right)+...+99\left(99+2\right)\)

\(=\left(1^2+3^2+5^2+...+99^2\right)+2\left(1+3+5+...+99\right)\)

Ta có:

\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)

⇒ \(A=\left(1^2+2^2+3^2+...+100^2\right)-2^2\left(1^2+2^2+3^2+...+50^2\right)+2\left(1+3+5+...+99\right)\)

\(=\dfrac{100.101.201}{6}+\dfrac{4.50.51.101}{6}+\dfrac{\left(99+1\right).\left[\left(99-1\right):2+1\right]}{2}\)

\(=338350-171700+5000\)

\(=166650+5000=171650\)

A=1.3+3.5+5.7+...+99.101

A=101

A = 171650 bạn nhé!