`Tính:` `A=(1-1/(1+2))(1-1/(1+2+3))(1-1/(1+2+3+4))(1-1/(1+2+3+5))+…….+(1-1/(1+2+3+…..+2015))` 04/11/2021 Bởi Jasmine `Tính:` `A=(1-1/(1+2))(1-1/(1+2+3))(1-1/(1+2+3+4))(1-1/(1+2+3+5))+…….+(1-1/(1+2+3+…..+2015))`
Đáp án: dangthanhhuong 28/01/2021 A=…=(1-12.3212.32)(1-13.4213.42)(1-14.5214.52) …. (1-12015.2016212015.20162) A=(1-22.322.3)(1-23.423.4)(1-24.524.5)…(1- 22015.201622015.2016) (1-12015.2016212015.20162) A=42.342.3 .103.4103.4 .184.5184.5…. 40622382015.201640622382015.2016 A=1.42.31.42.3 .2.53.42.53.4 .3.64.53.64.5…. 2014.20172015.20162014.20172015.2016 A=(1.2.3….2014)(4.5.6…..2017)(1.2.3….2015)(3.4.5….2016)(1.2.3….2014)(4.5.6…..2017)(1.2.3….2015)(3.4.5….2016) A=20172015.320172015.3 = 2017604520176045 vậy….. Cho mk xin ctlhn nha, thanks Giải thích các bước giải: Bình luận
A=…=(1-$\frac{1}{\frac{2.3}{2}}$)(1-$\frac{1}{\frac{3.4}{2}}$)(1-$\frac{1}{\frac{4.5}{2}}$) …. (1-$\frac{1}{\frac{2015.2016}{2}}$) A=(1-$\frac{2}{2.3}$)(1-$\frac{2}{3.4}$)(1-$\frac{2}{4.5}$)…(1- $\frac{2}{2015.2016}$) (1-$\frac{1}{\frac{2015.2016}{2}}$) A=$\frac{4}{2.3}$ .$\frac{10}{3.4}$ .$\frac{18}{4.5}$…. $\frac{4062238}{2015.2016}$ A=$\frac{1.4}{2.3}$ .$\frac{2.5}{3.4}$ .$\frac{3.6}{4.5}$…. $\frac{2014.2017}{2015.2016}$ A=$\frac{(1.2.3….2014)(4.5.6…..2017)}{(1.2.3….2015)(3.4.5….2016)}$ A=$\frac{2017}{2015.3}$ = $\frac{2017}{6045}$ vậy….. Cho mk xin ctlhn nha, thanks Bình luận
Đáp án:
A=…=(1-12.3212.32)(1-13.4213.42)(1-14.5214.52) …. (1-12015.2016212015.20162)
A=(1-22.322.3)(1-23.423.4)(1-24.524.5)…(1- 22015.201622015.2016) (1-12015.2016212015.20162)
A=42.342.3 .103.4103.4 .184.5184.5…. 40622382015.201640622382015.2016
A=1.42.31.42.3 .2.53.42.53.4 .3.64.53.64.5…. 2014.20172015.20162014.20172015.2016
A=(1.2.3….2014)(4.5.6…..2017)(1.2.3….2015)(3.4.5….2016)(1.2.3….2014)(4.5.6…..2017)(1.2.3….2015)(3.4.5….2016)
A=20172015.320172015.3 = 2017604520176045
vậy…..
Cho mk xin ctlhn nha, thanks
Giải thích các bước giải:
A=…=(1-$\frac{1}{\frac{2.3}{2}}$)(1-$\frac{1}{\frac{3.4}{2}}$)(1-$\frac{1}{\frac{4.5}{2}}$) …. (1-$\frac{1}{\frac{2015.2016}{2}}$)
A=(1-$\frac{2}{2.3}$)(1-$\frac{2}{3.4}$)(1-$\frac{2}{4.5}$)…(1- $\frac{2}{2015.2016}$) (1-$\frac{1}{\frac{2015.2016}{2}}$)
A=$\frac{4}{2.3}$ .$\frac{10}{3.4}$ .$\frac{18}{4.5}$…. $\frac{4062238}{2015.2016}$
A=$\frac{1.4}{2.3}$ .$\frac{2.5}{3.4}$ .$\frac{3.6}{4.5}$…. $\frac{2014.2017}{2015.2016}$
A=$\frac{(1.2.3….2014)(4.5.6…..2017)}{(1.2.3….2015)(3.4.5….2016)}$
A=$\frac{2017}{2015.3}$ = $\frac{2017}{6045}$
vậy…..
Cho mk xin ctlhn nha, thanks