1 + 1 =
Tìm số tự nhiên X biết 1_1/5 * 1_1/6 * 1_1/7 * …*1_1/2014 = X
1 +1_1+1_1 = ?
1+1-1+1-1
=2-1+1-1
=1+1-1
=2-1
=1
k mk nhe
A=1_1/3 x 1_1/8 x 1_1/15 x...x 1_1/399
B=1_4/5 x 1_4/21 x 1_4/45 x...x 1_4/437
ahhh khó quá
bạn nào giúp mình với
ai lam duoc cau nay minh cho 1 like
1_1+1_1+1
\(\dfrac{ }{ }\)1
_
1
1_1=?
1_1 bằng bao nhiêu
1 - 1 = 0
tk nha cảm ownnnnnnnnnnnnnnnnn !!!! ^^
(1_1/10)×(1-1/11)×(1-1/12)×(1-1×1/13)
(1-1/10)x(1-1/11)x(1-1/12)x(1-1x1/13)=\(\dfrac{9}{10}\)x\(\dfrac{10}{11}\)x\(\dfrac{11}{12}\)x\(\dfrac{12}{13}\)=\(\dfrac{9}{13}\).
1/2004×(1-1/2005)×(1_1/2006)×(1-1/2007)-(1-1/2008)
Ta có \(\frac{1}{2004}.\left(1-\frac{1}{2005}\right).\left(1-\frac{1}{2006}\right).\left(1-\frac{1}{2007}\right).\left(1-\frac{1}{2008}\right)\)
\(=\frac{1}{2004}.\frac{2004}{2005}.\frac{2005}{.2006}.\frac{2006}{2007}.\frac{2007}{2008}\)
\(=\frac{1.2004.2005.2006.2007}{2004.2005.2006.2007.2008}\)
\(=\frac{1}{2008}\)
\(\frac{1}{2004}\cdot\left(1-\frac{1}{2005}\right)\cdot\left(1-\frac{1}{2006}\right)\cdot\left(1-\frac{1}{2007}\right)\cdot\left(1-\frac{1}{2008}\right)\)
\(=\frac{1}{2004}\cdot\frac{2004}{2005}\cdot\frac{2005}{2006}\cdot\frac{2006}{2007}\cdot\frac{2007}{2008}\)
\(=\frac{1\cdot2004\cdot2005\cdot2006\cdot2007}{2004\cdot2005\cdot2006\cdot2007\cdot2008}=\frac{1}{2008}\)
1+1+1_1+1+2+7+7+8+9+8+7+7+6+8+6+9+6+9+6+86+8
1+1+1-1+1+2+7+7+8+9+8+7+7+6+8+6+9+6+9+6+86+8=106+86
=192
nhớ k cho mình nha