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\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)\)
\(=\dfrac{1}{3}.\dfrac{3n}{6n+4}\)
\(=\dfrac{n}{6n+4}\) ( đpcm )
Vậy...
Bài 4:
=>(x-5)*3/10=1/5x+5
=>3/10x-3/2=1/5x+5
=>1/10x=5+3/2=6,5
=>0,1x=6,5
=>x=65
a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1.2.3.4.5.6...39.40}{\left(2.4.6...40\right).21.22.23...40}=\frac{1.2.3.4.5.6...39.40}{2^{20}.1.2.3...20.21.22.23...40}\)
\(=\frac{1}{2^{20}}\left(đpcm\right)\)
b) \(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{\left(2.4.6...2n\right)\left(n+1\right)\left(n+2\right)...2n}=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{2^n.1.2.3...n\left(n+1\right)\left(n+2\right)...2n}\)
\(=\frac{1}{2^n}\left(đpcm\right)\)
Ta có:
\(\frac{1}{n}-\frac{1}{n+2}=\frac{n+2}{n\left(n+2\right)}-\frac{n}{n\left(n+2\right)}=\frac{n+2-n}{n\left(n+2\right)}=\frac{2}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{n\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)