a)\(=\dfrac{211}{180}\)

b)\(=\dfrac{5}{39}\)

c)=\(=-\dfrac{65}{168}\)

Cầu cứu mọi người hãy giúp đỡ mình

=>x(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+99/x=-3/7

=>8/9x+99/x=-3/7

\(\Leftrightarrow\dfrac{8x}{9}+\dfrac{99}{x}=\dfrac{-3}{7}\)

\(\Leftrightarrow\dfrac{8x^2+99\cdot9}{9x}=\dfrac{-3}{7}\)

\(\Leftrightarrow-56x^2-6237=27x\)

hay \(x\in\varnothing\)

a) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)

\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)

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

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2}.\dfrac{4}{15}=\dfrac{2}{15}\)

b) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)

\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}\left(-7\right)\)

\(=-7\left(\dfrac{4}{9}+\dfrac{59}{9}\right)\)

\(=-7.7=-49\)

c) \(\left(3\dfrac{2}{5}-2\dfrac{2}{5}\right).\left(-\dfrac{5}{3}\right)+3.\left(2\dfrac{1}{2}:\dfrac{1}{2}\right)\)

\(=\left(\dfrac{17}{5}-\dfrac{12}{5}\right).\left(-\dfrac{5}{3}\right)+3.5\)

\(=-\dfrac{5}{3}+15=13\dfrac{1}{3}\)

d) \(1\dfrac{13}{5}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)

\(=\dfrac{2}{7}+78\dfrac{8}{15}:\dfrac{47}{24}\)

( bạn tự tính nốt câu này nha ! )

\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)

\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)

\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)

\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)

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

\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)

\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)

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

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\) \(=\dfrac{5}{39}\)

\(\dfrac{5}{39}\)

=\(\dfrac{5}{39}\)

\(2A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}=\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=\dfrac{1}{3}-\dfrac{1}{13}=\dfrac{13-3}{39}=\dfrac{10}{39}\)

\(\Rightarrow A=\dfrac{10}{39}.\dfrac{1}{2}=\dfrac{10}{78}=\dfrac{5}{39}\)

Giải:

\(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\) 

\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\) 

\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\) 

\(=\dfrac{1}{1}-\dfrac{1}{13}\) 

\(=\dfrac{12}{13}\) 

Chúc em học tốt!

2/3+2/15+2/35+2/63+2/99+2/143

=2(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)

=2(1-1/3+1/3-1/5+1/5-....+1/13)

=2(1-1/13)

=2.12/13=24/13

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=1/2*10/39

=5/39

2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)