Tìm 3 chữ số tận cùng của :
\(A=1\cdot3\cdot5\cdot7\cdot....\cdot2013\cdot2015\)
Tìm 3 chữ số tận cùng ( BA CHỮ SỐ TẬN CÙNG) của
\(A=1\cdot3\cdot5\cdot7\cdot....\cdot2013\cdot2015\)
chứng minh rằng \(1\cdot3\cdot5\cdot....\cdot2013\cdot2015+2\cdot4\cdot6\cdot....\cdot2014\cdot2016⋮9911\)
Ta có:
\(9911=11\cdot17\cdot53\)
Để \(A=1.3.5...2015+2.4.6....2016⋮9911\)thì:\(\hept{\begin{cases}1.3.5...2015⋮9911\\2.4.6...2016⋮9911\end{cases}}\)
Mà: \(1.3.5...2015=1.3.5...11.13.15.17...53...2015⋮11.17.53=9911\)
và \(2\cdot4\cdot...\cdot2016=2\cdot4\cdot...\cdot22\cdot...\cdot34\cdot...\cdot106\cdot...\cdot2016⋮11\cdot17\cdot54=9911\)
=> đpcm
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
A = \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot.....\cdot\left(1+\frac{1}{2011\cdot2013}\right)\)
CHỨNG MINH RẰNG\(\frac{51}{2}\cdot\frac{52}{2}\cdot...\cdot\frac{100}{2}=1\cdot3\cdot5\cdot7\cdot...\cdot99\)
TA CÓ:\(1\cdot3\cdot....\cdot99=\frac{\left(1\cdot3\cdot...\cdot99\right)\left(2\cdot4\cdot...\cdot100\right)}{2\cdot4....\cdot100}=\frac{1\cdot2\cdot3\cdot....\cdot100}{2\cdot2\cdot2\cdot...\cdot2\left(50\right)\cdot1\cdot2\cdot3\cdot..\cdot50}\)
\(=\frac{51\cdot52\cdot...\cdot100}{2\cdot2\cdot2\cdot...\cdot2}=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}\)(ĐPCM)
\(A=\frac{2}{1\cdot3\cdot5}\cdot\frac{2}{5\cdot7\cdot9}\cdot...\cdot\frac{2}{97\cdot99\cdot101}\)
tính
\(E=\frac{2}{1\cdot3}\cdot\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(E=\frac{1}{1}-\frac{1}{99}\)
\(E=\frac{98}{99}\)
E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt
\(t\text{ìm}x\left(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+.....+\frac{36}{25\cdot27\cdot29}\right)\cdot x=\frac{4}{25}\)
\(36.\left(\frac{1}{1}-\frac{1}{3}-\frac{1}{5}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}-\frac{1}{29}\right).x=\frac{4}{25}\)
Triệt tiêu còn
\(36.\left(\frac{1}{1}-\frac{1}{29}\right).x=\frac{4}{25}\)
từ đây dễ rồi, tình lần lượt rồi tìm x nhé
Cho A=\(1\cdot3\cdot5\cdot7\cdot...\cdot49\)
B=\(\dfrac{26}{2}\cdot\dfrac{27}{2}\cdot...\cdot\dfrac{50}{2}\)
So sanhs A với B