1,Tính
(1-1/99)*(1-1/100)*...*(1-1/2005)
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\(\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right).....\left(1-\frac{1}{2005}\right)\)
\(=\frac{98}{99}.\frac{99}{100}.....\frac{2004}{2005}=\frac{98}{2005}\)
\(=\dfrac{98}{99}\cdot\dfrac{99}{100}\cdot...\cdot\dfrac{2004}{2005}\\ =\dfrac{98}{2005}\)
Ơ dễ mà =)
a, (1 + 3 + 5 + 7+ ... + 2003 + 2005) x (125125 x 127 - 127127 x125)
= (1 + 3 + 5 +...+2003 + 2005) x (125 x 1001 x 127 - 127x1001 x 125)
= (1 + 3 + 5 +...+ 2003 + 2005) x 0
= 0
\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{98}{100}=\frac{49}{50}\)
đề :
= 1/100 - (1 / 100.99 +1/99.98 + ...+ 1/3.2 +1/2.1 )
=1/100 - (1 /1.2 +1/ 2.3 +...+ 1/ 98.99 +1 / 99.100)
=1/100 -( 1- 1/ 2 +1/2 -1/3 +...+1/98 -1/99 +1/99 -1/100)
=1/100 - ( 1- 1/100)
=1/100 - 99 /100
= -98/100
= -49 /50
Ta có \(63,1.2-21,3.6=0,9.7.10.1,2-21.3,6\)
\(=6,3.1,2-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.1,2-6,3.1,2\)
\(=0\)
\(\Rightarrow\dfrac{\left(1+2+......+100\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}=\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+......+99-100}=0\)
(1-1/99)*(1-1/100)*...*(1-1/2005)
\(=\frac{98}{99}\cdot\frac{99}{100}\cdot...\cdot\frac{2004}{2005}\)
\(=\frac{98\cdot99\cdot...\cdot2004}{99\cdot100\cdot...\cdot2005}\)
\(=\frac{98}{2005}\)
Ta co:1-1/99=98/99;
1-1/100=99/100
…
1- 1/2005=2004/2005
Nen h tren = 98/99×99/100x…..x2004/2005= 98/2005