\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)

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

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

\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

Mk sửa lại 1 tí nha dòng thứ 5 :

\(A=6-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}\left(1-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}.\frac{12}{13}\)

\(=6-\frac{6}{13}=\frac{72}{13}\)

Mong bn bỏ qua nha

Sửa đề: \(98+99+\dfrac{142}{144}\) \(\rightarrow\dfrac{98}{99}+\dfrac{143}{144}\)  

Giải:

\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+\dfrac{62}{63}+\dfrac{98}{99}+\dfrac{143}{144}+\dfrac{194}{195}\) 

\(A=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{195}\right)\) 

\(A=7-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{195}\right)\) 

\(A=7-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\dfrac{14}{15}\right]\) 

\(A=7-\dfrac{7}{15}\) 

\(A=\dfrac{98}{15}\) 

Chúc bạn học tốt!

\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

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

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

\(=5+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(=5+\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(=5+\frac{5}{11}=\frac{60}{11}\)

A = \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

A = ( 1 - 1/3 ) + ( 1 - 1/15 ) + ( 1 - 1/35 ) + ( 1 - 1/63 ) + ( 1 - 1/99 )

A = ( 1 + 1 + 1 + 1 + 1 ) - ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )

A = 5 - \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

A  = 5 - ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 )

A = 5 - ( 1 - 1/11 ) 

A = 5 - 10/11

A = 45/11

Dấu \(.\)là dấu nhân 

\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(\Rightarrow A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)

\(\Rightarrow A=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(\Rightarrow A=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(\Rightarrow A=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(\Rightarrow A=5-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

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

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

\(\Rightarrow A=5-\frac{5}{11}\)

\(\Rightarrow A=\frac{55}{11}-\frac{5}{11}\)

\(\Rightarrow A=\frac{50}{11}\)

~ Ủng hộ nhé 

= 50/11 nha

Chúc bạn hok tốt nhé!

Dấu chấm là dấu nhân

\(P=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(P=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)

\(P=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(P=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(P=5-\frac{1}{2}.2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(P=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(P=5-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(P=5-\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

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

\(P=5-\frac{5}{11}\)

\(P=\frac{55}{11}-\frac{5}{11}\)

\(P=\frac{50}{11}\)

+ \(\frac{1}{n\times\left(n+2\right)}=\frac{\left(n+2\right)-n}{n\times\left(n+2\right)}\)

\(=\frac{n+2}{n\times\left(n+2\right)}-\frac{n}{n\times\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)

+ \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

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

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

\(=5-\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)

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

\(=5-\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)

\(=5-\frac{1}{2}+\frac{1}{22}=\frac{50}{11}\)

                              =50/11

       =50/11

#)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:

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

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

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

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

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

\(\Leftrightarrow2011-12x=91\)

\(\Leftrightarrow12x=1920\)

\(\Leftrightarrow x=160\)