\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)

\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)

\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)

nhìn công thức đây này \(\sqrt[]{\sqrt{ }\hept{\begin{cases}\\\end{cases}}\frac{ }{ }^{ }_{ }\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}}\) xong rồi đó không cần cảm ơn

(1/15+1/35+1/63+1/99).x=4

4/33.x=4

x=4:4/33

x=16/33

Ta có \(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).x=4\)

=> \(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=4\)

=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=4\)

=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).x=8\)

=> \(\left(\frac{1}{3}-\frac{1}{11}\right).x=8\)

=> \(\frac{8}{33}.x=8\)

=> x = 33

Vậy x = 33

#)Giải :

\(200-18:\left(372:3x-1\right)-28=166\)

\(\Leftrightarrow200-18:\left(372:3x-1\right)=194\)

\(\Leftrightarrow18:\left(372:3x-1\right)=6\)

\(\Leftrightarrow372:3x-1=3\)

\(\Leftrightarrow3x-1=124\)

\(\Leftrightarrow3x=125\)

\(\Leftrightarrow x=\frac{125}{3}\)

200 - 18 : (372 : 3 . x - 1) - 28 = 166

=> 200 - 18 : (372 : 3.x - 1)     = 166 + 28

=> 200 - 18 : (372 : 3.x) - 1)    = 194

=>          18 : (372 : 3.x - 1)     = 200 - 194

=>          18 : (372 : 3.x - 1)     = 6

=>                  372 : 3.x  - 1      = 18 : 6

=>                  372 : 3.x - 1       = 3

=>                    372 : 3.x          = 3 + 1

=>                    372 : 3.x          = 4

=>                             3.x          = 372 : 4

=>                             3.x          = 93

=>                                x          = 93 : 3

=>                                x          = 31

#)Next:

\(\left(100-99+98-97+...+2-1\right):50+2010-12x=91\)

\(\Leftrightarrow\left[\left(100-99\right)+\left(98-97\right)+...+\left(2-1\right)\right]:50+2010-12x=91\)

\(\Leftrightarrow\left[1+1+...+1\right]:50+2010-12x=91\)

\(\Leftrightarrow50:50+2010-12x=91\)

\(\Leftrightarrow1+2010-12x=91\)

\(\Leftrightarrow2011-12x=91\)

\(\Leftrightarrow12x=1920\)

\(\Leftrightarrow x=160\)

cái này tính cái gì thế

ko hiểu

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 \(x=\frac{22}{15}\)