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 

còn cần không bạn, mk làm cho

\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)

\(\frac{99}{20}-2x=\frac{49}{99}\)

\(2x=\frac{99}{20}-\frac{49}{99}\)

\(2x=\frac{8821}{1980}\)

\(x=\frac{8821}{1980}:2\)

\(x=\frac{8821}{3960}\)

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!

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

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)

\(\dfrac{1.2}{2.1.3}+\dfrac{1.2}{2.3.5}+\dfrac{1.2}{2.5.7}+...+\dfrac{1.2}{2x\left(x+2\right)}\dfrac{5}{11}\)\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{5}{11}\)\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)\(\dfrac{1}{2}\left(1+0+0+...+0+\dfrac{1}{x-2}\right)=\dfrac{5}{11}\)

\(\dfrac{1}{2}\left(1-\dfrac{1}{x-2}\right)=\dfrac{5}{11}\)

\(1-\dfrac{1}{x-2}=\dfrac{5}{11}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x-2}=\dfrac{10}{11}\)

\(\dfrac{1}{x-2}=1-\dfrac{10}{11}\)

\(\dfrac{1}{x-2}=\dfrac{1}{11}\)

\(\Rightarrow x-2=11\)

\(x=11+2\)

\(x=13\)

Vậy x=13

 nhung ma ko cothoi gian giai

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

làm tắt thế ai mà hỉu đc

hỏi gì nhiều thế

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