\(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{8}{3}\)

hay \(x=\dfrac{4}{9}\cdot\dfrac{3}{8}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)

Có: \(x=\dfrac{1}{9}+\dfrac{8}{116}=\dfrac{1}{9}+\dfrac{2}{29}=\dfrac{47}{261}\)

\(x-\left(\dfrac{2+2+2+2}{3+15+35+63}\right)=\dfrac{1}{9}\)

\(\Leftrightarrow x=\dfrac{1}{9}+\dfrac{2}{29}=\dfrac{47}{261}\)

a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)

b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)

\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)

\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)

\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)

\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)

Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{9.11}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

Thay A vào biểu thức

\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)

\(\Rightarrow x=\frac{22}{15}\)

P/s: Có thể tính sai :(

\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)

Trước tiên mình tính dãy có dấu ngoặc đã

Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)

\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)

\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)

\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)

Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :

\(\frac{5}{11}\times x=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)

\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)

\(\Leftrightarrow x=\frac{22}{15}\)

Vậy \(x=\frac{22}{15}\)

(2/5+2/35+2/63):x=1/18

       2/9           :x=1/18

                        x=2/9:1/18

                        x=4

cái này tính cái gì thế

ko hiểu

a) CÓ: A = (1-1/4²).(1-1/5²).(1-1/6²)......(1-1/200²)

               =\(\frac{4^2-1^2}{4^2}\). \(\frac{5^2-1^2}{5^2}\). \(\frac{6^2-1^2}{6^2}\)....... \(\frac{200^2-1^2}{200^2}\)

Ta có công thức sau : a²-b²= a² -ab+ab-b² 

                                            = a(a-b) + b(a-b)

                                            = (a+b)(a-b)

   ÁP DỤNG CÔNG THỨC TRÊN VÀO BÀI TOÁN TA ĐƯỢC : 

  A=  \(\frac{3.5}{4^2}\). \(\frac{4.6}{5^2}\). \(\frac{5.7}{6^2}\)......\(\frac{199.201}{200^2}\)

    = \(\frac{\left(3.4.5.....199\right)\left(5.6.7....201\right)}{\left(4.5.6......200\right)^2}\)

    =    \(\frac{\left(3.4.5.......199\right)\left(5.6.7.....200.201\right)}{\left(4.5.6.....199.200\right)\left(4.5.6......200\right)}\)

    =   \(\frac{3.201}{200.4}\)

   =  \(\frac{603}{800}\)​​

b)Từ đề bài ta suy ra : B=\(\frac{1.3}{5.7}\).\(\frac{3.5}{7.9}\). \(\frac{5.7}{9.11}\)...... \(\frac{99.101}{103.105}\)

                                      = \(\frac{1.3^2.5^2.7^2......99^2.101}{5.7^2.9^2.11^2....99^2.101^2.103^2.105}\)

                                      =\(\frac{3^2.5}{101.103^2.105}\)

                                       =\(\frac{3}{7500563}\)

Tk mình đi mọi người mình bị âm nè!

ai tk mình mình tk lại cho!!!

( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 ) x X = 2/3

                                                       X = 2/3 : ( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 )

                                                       X = 286/85

k mình đi mình đang bị âm

x=1,466666667

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

\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)

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

\(B=\frac{1}{3}-\frac{1}{21}\)

\(B=\frac{2}{7}\)

A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\) 

A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)

A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)

A=\(\frac{1}{1}-\frac{1}{11}\)

=>A=\(\frac{10}{11}\)

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

2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

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

2B=\(\frac{2}{7}\)

B=\(\frac{2}{7}:2\)

=>B=\(\frac{1}{7}\)