K
Khách

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

\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{99.101}\right)\)

\(=\frac{2.2}{1.3}\frac{3.3}{2.4}.....\frac{100.100}{99.101}\)

\(=\frac{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}\)

\(=\frac{100.2}{101}=\frac{200}{101}\)

\(\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right)^2}< 0\)

\(\Rightarrow\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right).\left(x-2\right)}< 0\)

=> ( x - 3 ) . ( x - 5 ) và ( x - 2 ) . ( x - 2 ) trái dấu 

Mà ( x - 2 )2 = ( x - 2 ) . ( x - 2 ) ≥ 0 ∀ x

 \(\Rightarrow\hept{\begin{cases}\left(x−3\right).\left(x+5\right)< 0\\\left(x−2\right).\left(x−2\right)>0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< −5;−5< x< 3\\x>2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< −5\\2< x< 3\end{cases}}\)

17 tháng 4 2019

2

\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\frac{25}{51}\)

\(S1=\frac{25}{102}\)

14 tháng 10 2023

\(M=\dfrac{2^2.3^2.4^2.....20^2}{1.3.2.4.3.5.4.6.5.7.6.8.7.9....19.21}=\)

\(=\dfrac{2^2.3^2.4^2....20^2}{1.2.3^2.4^2....19^2.20.21}=\dfrac{2.20}{21}=\dfrac{40}{21}\)

\(N=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{10}{11}=\dfrac{1}{11}\)