Bài 4: \(C=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
\(=7\left(\frac{5}{2\cdot7}+\frac{4}{7\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\right)\)
\(=7\left(\frac12-\frac17+\frac17-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=7\left(\frac12-\frac{1}{28}\right)=7\cdot\frac{14-1}{28}=\frac{13}{4}\)
`C = 5/(2*1) + 4/(1*11) + 3/(11*2) +1/(2*15) + 13/(15*4)`
`=> C/7 = 5(2*7) + 4/(7*11) + 3/(11*14) + 1/(14*15) + 13/(15*28)`
`=> C/7 = 1/2 - 1/7 + 1/7 - .... -1/28`
`=> C/7 = 1/2 -1/28`
`=>C/7 = 13/28`
`=> C = 13/4`
giúp mình làm bài 4 nha ai trả lời đúng là trai xinh gái đẹp nha,lớp 7 học sinh xuất sắc luôn
bài 1:
D=\(\frac{3^2}{2\cdot5}+\frac{3^2}{5\cdot8}+\frac{3^2}{8\cdot11}+\frac{3^2}{11\cdot14}+\frac{3^2}{14\cdot17}\)
\(D=3\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}\right)\)
áp dụng công thức: \(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
=> \(D=3\left(\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
\(D=3\left(\frac12-\frac{1}{17}\right)\)
\(D=3\cdot\frac{15}{34}\)
\(D=\frac{45}{34}\)
Bài 2:
=> 2C=\(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{98\cdot99\cdot100}\)
áp dụng công thức: \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(2C=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\cdots+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
=> \(2C=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)
\(2C=\frac{4949}{9900}\)
=> \(C=\frac{4949}{19800}\)
Bài 3:
\(\Rightarrow3B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{1}{8\cdot11}+\ldots+\frac{3}{2015\cdot2018}\)
\(3B=\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}+\cdots+\frac{1}{2015}-\frac{1}{2018}\)
\(3B=\frac{504}{1009}\)
=> \(B=\frac{168}{1009}\)
Bài 5:
\(C=-\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(C=-\left(\frac14-\frac15+\frac15-\frac16+\frac16-\frac17+\frac17-\frac18+\frac18-\frac19+\frac19-\frac{1}{10}\right)\)
\(C=-\left(\frac14-\frac{1}{10}\right)\)
\(C=-\frac{3}{20}=-0,15\)
Bài 6:
=> \(B=\frac23\left(\frac{3}{60\cdot63}+\frac{3}{63\cdot66}+\frac{3}{66\cdot69}+\cdots+\frac{3}{117\cdot120}\right)\)
=> \(B=\frac23\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+\frac{1}{66}-\frac{1}{69}+\cdots+\frac{1}{117}-\frac{1}{120}\right)\)
\(B=\frac23\left(\frac{1}{60}-\frac{1}{120}\right)\)
\(B=\frac{1}{180}\)
bài 7:
\(A=2019-\frac{1}{2019}-2018+\frac{1}{2019}\)
\(A=1\)
bài 8:
\(a\cdot B=-\left(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot\ldots\cdot\frac{399}{20^2}\right)\)
\(a\cdot B=-\frac{\left(1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\ldots\cdot19\cdot21\right)}{\left(2\cdot2\right)\cdot\left(3\cdot3\right)\cdot\left(4\cdot4\right)\cdot\ldots\cdot\left(20\cdot20\right)}\)
\(a\cdot B=\frac{\left(1\cdot2\cdot3\cdot4\cdot\ldots\cdot19\right)\left(3\cdot4\cdot5\cdot\ldots\cdot21\right)}{\left(2\cdot3\cdot4\cdot\ldots\cdot20\right)\left(2\cdot3\cdot4\cdot\ldots\cdot20\right)}\)
\(a\cdot B=-\left(\frac{1}{20}\cdot\frac{21}{2}\right)\)
\(a\cdot B=-\frac{21}{40}=-0,525\)
