Tính nhanh :
( 1 - 1/99 ) x ( 1- 1/100 ) x ... x ( 1- 2005 )
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M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)
M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
M = \(1-\dfrac{1}{100}\)
M = \(\dfrac{99}{100}\)
\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times...\times1\frac{1}{100}\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times...\times\frac{101}{100}\)
\(=\frac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)
\(=\frac{101}{2}\)
\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right)+\left(1+\frac{1}{98}\right)+....+\left(1+\frac{1}{2}\right)\)
= \(\frac{101}{100}.\frac{100}{99}.\frac{99}{98}.....\frac{3}{2}\)
= \(\frac{100}{2}\)
= 50
a)
Ta có : ( 1 + 2 + 3 + ... + 99)
Số số hạng là: ( 99 - 1 ) : 1 + 1 = 100
Tổng là: ( 99 + 1 ) x 100 : 2 = 5000
=> 5000 x ( 13 - 12 - 1 ) x 15
=> 5000 x 10 x 15
=> 50000 x 15
=> 750000
Ko muốn vt nx :))
Xin phép sửa lại đề.
\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)\)
\(1-\frac{1}{2}=\frac{1}{2}\)
\(1-\frac{1}{3}=\frac{2}{3}\)
\(1-\frac{1}{4}=\frac{3}{4}\)
.........................
\(1-\frac{1}{99}=\frac{98}{99}\)
\(1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow P=\frac{1.2.3...98.99}{2.3.4...99.10}\)
\(P=\frac{\left(1.2.3...98.99\right)}{\left(2.3.4...99.100\right)}\)
\(P=\frac{1}{100}\)
Vậy: P = 1/100
bài 1"
121/27 x 54/11 < n < 100/21 : 25/126
=> 22 < n < 24
mà n là số tự nhiên
=> n = 23
Bài 2:
(m:1 - m x 1) : ( m x 2005 + m + 1)
= (m-m) : ( m x 2005 + m + 1)
= 0 : (m x 2005 + m + 1) = 0
1.
\(\frac{121}{27}\times\frac{54}{11}< n< \frac{100}{21}:\frac{25}{126}\)
=> \(22< n< 24\)
Mà n là số tự nhiên
=> n = 23
2.
( m : 1 + m x 1 ) : ( m x 2005 + m + 1 )
= ( m - m ) : ( m x 2005 + m + 1 )
= 0 : ( m x 2005 + m + 1 )
= 0
\(\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right).....\left(1-\frac{1}{2005}\right)\)
\(=\frac{98}{99}.\frac{99}{100}.....\frac{2004}{2005}=\frac{98}{2005}\)