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

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 \(x=\frac{22}{15}\)

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

câu c hình như sai đề hả bn

A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{129.15}\)

A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{129.15}\)

A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{129}-\frac{1}{15}\)

A = \(\frac{1}{3}-\frac{1}{15}\)= \(\frac{5}{15}-\frac{1}{15}\)

A = \(\frac{4}{15}\)

CẢM ƠN LỜI GỢI Ý

kết quả đúng là 7/45 nhé

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

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

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

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

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

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

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

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

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

                              =50/11

       =50/11

giúp tớ với

Câu 1: 

a) \(=\frac{42}{5}:\frac{14}{5}\)

\(=3\)

b)

 \(=15-\frac{7}{3}:\left(\frac{4}{9}-\frac{1}{6}\right)\)

\(=15-\frac{7}{3}:\frac{4}{9}+\frac{7}{3}:\frac{1}{6}\)

\(=15-\frac{21}{4}+14\)

\(=29-\frac{21}{4}=29-5-\frac{1}{4}\)

\(=24-\frac{1}{4}=\frac{95}{4}\)

tách ra đi 

Bài 1 : Tính

a) (-14) + (-28) = -42

b) (-26) + 47 = 21

c) 35 + (-63) = -28

d) (-8) + 12 + (-24) = -(8 + 24) + 12 =  -32 + 12 = -20

e) (-26) - 27 = -53

f) (-35) - ( 105) [ Tôi không biết làm :v ]

Bài 2 :

a) 54 + (12 + 46) = 54 + 12 + 46 = 54 + 46 + 12 = 100 + 12 = 112

b) 157 - ( 200 + 57 ) = 157 - 200 + 57 = 157 - 57 - 200 = -100

Còn lại tự tính :v

mik chỉ làm bài 1 thôi nha  bài bạn gửi dài quá

^{_{Bài 1 : Tính}}

^{_{a) (-14) + (-28) = -42}}

^{_{b) (-26) + 47 = 21}}

^{_{c) 35 + (-63) = -28}}

^{_{d) (-8) + 12 + (-24) = -(8 + 24) + 12 =  -32 + 12 = -20}}

^{_{e) (-26) - 27 = -53}}

^{_{f) (-35) - ( 105) =-140}}

Bài 1: Tính 

a) (−14) + (− 28) = -42                b) ( −26) + 47= 21          c) 35 + (− 63) = -28                                        

d) ( −8) +12 +( −24)= -20            e) ( −26)−27= -53           f) ( −35)− ( −105)=70  

Bài 2: Tính hợp lý 

a) 54 +(12+ 46)= 112          b) 157− (200+57) =-100                 

 c) (− 34) [(− 34)+2020]=(-34)+1986=1952   

d) 45− (137− 155)=63   e) 27− 8− 12− 7 =0                 

 f) ( −72)+26 −28− 6 =-80 

\(A=\left(\frac{1}{3}-\frac{8}{15}-\frac{1}{7}\right)+\left[\frac{2}{3}-\left(-\frac{7}{15}\right)+\frac{8}{7}\right]\)

\(A=\frac{1}{3}-\frac{8}{15}-\frac{1}{7}+\frac{2}{3}+\frac{7}{15}+\frac{8}{7}\)

\(A=\left(\frac{1}{3}+\frac{2}{3}\right)+\left(-\frac{8}{15}+\frac{7}{15}\right)+\left(-\frac{1}{7}+\frac{8}{7}\right)\)

\(A=1+\left(-\frac{1}{15}\right)+1\)

\(A=\frac{29}{15}\)

\(B=\left[\left(0,25+\frac{3}{5}\right)-\left(\frac{1}{8}-\frac{2}{5}+\frac{5}{4}\right)\right]\)

\(B=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-\frac{5}{4}\)

\(B=\left(\frac{1}{4}-\frac{5}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)-\frac{1}{8}\)

\(B=-1+1-\frac{1}{8}\)

\(B=0-\frac{1}{8}\)
\(B=-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{8}{15}-\frac{1}{7}+\frac{2}{3}+\frac{7}{15}+\frac{8}{7}\)

\(=\left(\frac{1}{3}+\frac{2}{3}\right)-\left(\frac{8}{15}+\frac{7}{15}\right)-\left(\frac{1}{7}-\frac{8}{7}\right)\)

\(=1-1+1\)

\(=0\)

\(A=\left(\frac{1}{3}-\frac{8}{15}-\frac{1}{7}\right)+\left(\frac{2}{3}-\left(-\frac{7}{15}\right)+\frac{8}{7}\right)\)

\(A=\frac{1}{3}-\frac{8}{15}-\frac{1}{7}+\frac{2}{3}+\frac{7}{15}+\frac{8}{7}\)

\(A=\left(\frac{1}{3}+\frac{2}{3}\right)-\left(\frac{8}{15}-\frac{7}{15}\right)+\left(\frac{8}{7}-\frac{1}{7}\right)\)

\(A=1-\frac{1}{15}+1\)

\(A=\frac{29}{15}\)

\(B=\left[0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+\frac{5}{4}\right)\right]\)

\(B=\left[\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-\frac{5}{4}\right]\)

\(B=\left(\frac{1}{4}-\frac{5}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)-\frac{1}{8}\)

\(B=\left(-1\right)+1-\frac{1}{8}\)

\(B=0-\frac{1}{8}\)

\(B=-\frac{1}{8}\)