Giải:

\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\) 

\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{19.21}=\dfrac{x}{49}\) 

\(2.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\) 

\(2.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=\dfrac{x}{49}\) 

\(2.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{x}{49}\) 

\(2.\dfrac{2}{7}=\dfrac{x}{49}\) 

   \(\dfrac{4}{7}=\dfrac{x}{49}\) 

\(\Rightarrow x=\dfrac{4.49}{7}=28\) 

Chúc bạn học tốt!

 \(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)

 2 . \(\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{399}=\dfrac{x}{49}\) 

 2 . \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{19.21}=\dfrac{x}{49}\)

 2 . ( \(\dfrac{1}{3}-\dfrac{1}{21}\) ) = \(\dfrac{x}{49}\)

 2 . \(\dfrac{2}{7}\) = \(\dfrac{x}{49}\)

=> \(\dfrac{4}{7}=\dfrac{x}{49}\)

=> \(\dfrac{21}{49}=\dfrac{x}{49}\)

=> \(x=21\)

Vậy \(x=21\)

Ta có : \(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)

\(\Leftrightarrow2\cdot\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{x}{98}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{98}\)

\(\Leftrightarrow\dfrac{2}{7}=\dfrac{x}{98}\Rightarrow x=28\)

Vậy $x=28$

a.\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}=\frac{37}{111}-\frac{1}{111}=\frac{36}{111}=\frac{12}{37}\)

Vậy A=\(\frac{12}{37}\)

b.\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)

Vậy \(B=\frac{2}{7}\)

c.\(C=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

\(\Rightarrow C.\frac{1}{2}=\left(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\right).\frac{1}{2}\)

\(=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)

\(=\frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}\)

Vậy \(C=\frac{3}{16}\)

A = \(\frac{4}{3.7}+\frac{4}{7.9}+...+\frac{4}{107.111}\)

A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}\)=\(\frac{12}{37}\)

2 câu sau tương tự. Mik ngại làm lắm -_-

\(A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(A=\frac{1}{3}-\frac{1}{111}\)

\(A=\frac{12}{37}\)

mà dài quá bạn ơi ban tách ra thành nhiều câu hỏi đi thế này trả lời lâu lắm

\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=4.\left(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{399}\right)=4.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\right)=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\right]=4.\left[\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\right]=2.\left(\dfrac{7-1}{21}\right)=\dfrac{12}{21}=\dfrac{4}{7}\)

dễ lm bạn ơi

B=4/3.5+4/5.7+.........................

Ghi tới đok chắc hỉu roài chứ, tự lm phần sau nhé.

bạn hỏcais gì vậy mình ko hiểu 

b= 4/3 + 4/15 + 4/35 + 4/63 + 4/99 + 4/143

là sao vậy bn

kết quả là

7/41

\(B=\frac{4}{3}+\frac{4}{15}+\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}\)

\(B=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+\frac{4}{11.13}\)

sorry,mình đi ngủ

\(B=\frac{4}{3}+\frac{4}{15}+\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{1}{143}\)

\(B=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+\frac{4}{11.13}\)

\(B\div2=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.11}+\frac{2}{11.13}\)

\(B\div2=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(B\div2=1-\frac{1}{13}=\frac{12}{13}\)

\(B=\frac{12}{13}.2=\frac{24}{13}\)