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đặt \(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{65.68}\right)\)là A
Ax=\(\frac{19}{68}+\frac{7}{34}=\frac{33}{68}\)
3A=\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{11}{8.11}+...+\frac{1}{65.68}\right)\)
3A=\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\)
3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\)
3A=\(\frac{1}{2}-\frac{1}{68}=\frac{33}{68}\)
A=33/68:3=11/68
\(\Rightarrow\)33/68:11/68=3
vậy x= 3
<=> \(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
<=>\(x+3=308\)
<=>\(x=305\)
Ta có : A = 1/ 2.5 + 1/ 5.8 + 1/ 8.11 + ... + 1/ (3n-1).(3n+2) .
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/ 3n-1 - 1/ 3n+2 .
= 1/2 - 1/ 3n+2 .
= 3n + 2 - 2 / 2 .( 3n+2 ) .
= 3n / 2.(3n+2) .
bai 1:\(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\right)x-\frac{7}{34}=\frac{19}{68}\)
=\(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)x-\frac{7}{34}=\frac{19}{68}\)
\(\Rightarrow\)x=3
bai2:từ giả thiết \(\frac{ab}{bc}=\frac{a}{c}=\frac{-1}{2}va\frac{ab}{ac}=\frac{b}{c}=\frac{3}{4}\)
hay \(\frac{a}{-2}=\frac{b}{3}=\frac{c}{4}\)
\(\Rightarrow\)\(\left(\frac{a}{-2}\right)^2=\frac{a}{-2}\times\frac{b}{3}=\frac{-6}{-6}=1\)
a=-2 (a<0)
\(\Rightarrow\)a=-2,b=3,c=4
Ta có:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right).\left(3n+5\right)}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3n+2\right).\left(3n+5\right)}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n+2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{6}-\frac{1}{9n+15}\)
(1/2.5+1/5.8+....+1/65.68)x=19/68+7/34=33/68
(1/2.5+1/5.8+.....+1/65.68).3.x=33/68.3=99/68
(3/2.5+3/5.8+........3/65.68)x =99/68
(1/2-1/68)x=99/68
33/68x=99/68
=>x=3
cho mình nha