1009/2016

\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)

\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)

\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)

\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)

101/12 bạn nha

CHÚC BẠN HỌC GIỎI

= 3/2 + 4/3 + 5/4  ................................ 100/99

= 100/2 = 50

\(\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right).....\left(1-\frac{1}{2006}\right)\)

\(=\left(\frac{99}{99}-\frac{1}{99}\right).\left(\frac{100}{100}-\frac{1}{100}\right).....\left(\frac{2006}{2006}-\frac{1}{2006}\right)\)

\(=\frac{98}{99}.\frac{99}{100}......\frac{2005}{2006}\)

\(=\frac{98.99.....2005}{99.100....2006}\)

\(=\frac{98}{2006}=\frac{49}{2006}\)

ủng hộ nha ai k mik k lại

a)  \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)

b)  \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)

c)  \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)

d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)

e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)

\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)

tìm số tự nhiên nhỏ nhất biết rằng khi số này cho 23 du 21 khi chia 17du 16

bạn đang trả lời giúp mình hay hỏi mình vậy

C = \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...............x\frac{1996}{1997}\)
C = \(\frac{1}{1997}\)

Câu b: Đặt  \(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2004}-1\right)\)

Ta có:  \(\frac{1}{2}-1=\left(-\frac{1}{2}\right);\frac{1}{3}-1=\left(-\frac{2}{3}\right);...;\frac{1}{2004}-1=\left(-\frac{2003}{2004}\right)\)

\(\Rightarrow B=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{2003}{2004}\right)\)

Vì B là 2003 thừa số âm nhân lại với nhau nên B là số âm

\(\Rightarrow B=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\right)=-\frac{1}{2004}\)

Câu a: Đặt  \(A=1+2^4+2^8;B=1+2+2^2+...+2^{11}\)

\(\Rightarrow16A=2^4+2^8+2^{12}\)   \(\Rightarrow15A=2^{12}-1\)   \(\Rightarrow A=\frac{2^{12}-1}{15}\)    \(\left(1\right)\)

\(\Rightarrow2B=2+2^2+2^3+...+2^{12}\)   \(\Rightarrow B=2^{12}-1\)   \(\left(2\right)\)

Từ  \(\left(1\right)\) và    \(\left(2\right)\)   \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)

biết làm bài 1 thôi

\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)

= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)

lượt bỏ đi còn :

\(\frac{1000}{2}=500\)