K
Khách

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

13 tháng 6 2020

\(b,\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{100}{2}\)

\(=50\)

8 tháng 3 2019

Mk ko biết lm nhưng cứ k thoải mái nha

SORRY

25 tháng 7 2017

A=   \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+\frac{4}{3}\right)\)

A=  \(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}-\frac{6}{13}+\frac{1}{2}+\frac{4}{3}\)

A=   \(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{7}{13}+\frac{6}{13}\right)-\left(\frac{1}{3}-\frac{4}{3}\right)\)

A=    \(1-1-\left(-1\right)\)

A=   \(1\)

B=   \(0,75+\frac{2}{5}+\left(\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\right)\)

B=    \(\frac{3}{4}+\frac{2}{5}+\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\)

B=    \(\left(\frac{3}{4}+\frac{5}{4}\right)+\left(\frac{2}{5}-\frac{7}{5}\right)+\frac{1}{9}\)

B=     \(2-1+\frac{1}{9}\)  

B=      \(\frac{9}{9}+\frac{1}{9}\)

B=    \(\frac{10}{9}\)

25 tháng 7 2017

C=   \(\left(\frac{-3}{2}.\frac{4}{3}\right).\left(\frac{-9}{2}\right)-\frac{1}{4}\)

C =   \(-2.\left(\frac{-9}{2}\right)-\frac{1}{4}\)

C =    \(9-\frac{1}{4}\)

C =   \(\frac{36}{4}-\frac{1}{4}\)

C =   \(\frac{35}{4}\)

D =   \(\frac{5}{4}.\left(\frac{-7}{10}.\frac{5}{4}-\frac{7}{8}.\frac{7}{10}\right)\)

D =    \(\frac{5}{4}.\left(\frac{-7}{8}-\frac{49}{80}\right)\)

D =     \(\frac{-35}{32}-\frac{49}{64}\)

D =     \(\frac{-70}{64}-\frac{49}{64}\)

D =     \(\frac{-119}{64}\)

k mk nha ^_^ 

A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)

A = \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{2004}{2004}-\frac{1}{2004}\right)\)

A = \(\frac{1}{2}\)x\(\frac{2}{3}.\)\(\frac{3}{4}....\)\(\frac{2003}{2004}\)

A = \(\frac{1}{2004}\)