K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 8 2017

\(15-\left\{2.\left[\left(2x-4\right).5\right].3.\left(x+1\right)\right\}=12-x\)

\(15-\left\{\left[10x-20\right].6.\left(x+1\right)\right\}=12-x\)

\(15-\left\{10x-20.6x+1\right\}=12-x\)

\(15-\left\{10x-120x+1\right\}=12-x\)

\(15-\left(-110x\right)-1=12-x\)

\(15+110x-1=12-x\)

\(110x+x=12-15+1\)

\(111x=-2\)

\(x=\dfrac{-2}{111}\)

19 tháng 7 2017

\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)

\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)

\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)

\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)

\(=\dfrac{3.2}{1.1}=6\)

24 tháng 7 2016

Ta có : 

\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)

\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)

\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)

\(=\frac{2015}{1}.\frac{2}{2016}\)

\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)

\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)

Vậy \(x=2015\)

Ủng hộ mk nha !!! ^_^

24 tháng 7 2016

ê cần giúp ko0

21 tháng 4 2023

Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).

Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\)\(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\)\(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\)\(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\) 

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)

\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)

10 tháng 1 2022

bằng 0 nha bạn

tick cho mình

10 tháng 1 2022

\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)

1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)

2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)