Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 :))
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
(125 x 4 - 1000 : 2) x ( 1 x2 x 3 x... x 99 x 100)
=(500 - 500 ) x ( 1 x2 x 3 x... x 99 x 100)
=0 x ( 1 x2 x 3 x... x 99 x 100)
=0
( 125 x 4 - 1000 : 2 ) x ( 1 x 2 x 3 x ... x 99 x 100 )
Đặt A = 125 x 4 - 1000 : 2
B = 1 x 2 x 3 x ... x 99 x 100
A = 125 x 4 - 1000 : 2
A = 500 - 500
A = 0
Thay vào , ta có :
0 x ( 1 x 2 x 3 x ... x 99 x 100 )
= 0
\(\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times..........\times\left(1+\frac{1}{2005}\right)\)
=\(\frac{3}{2}\times\frac{4}{3}\times............\times\frac{2006}{2005}\)
=\(\frac{2006}{2}=1003\)
a)\(=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times...\times\left(\dfrac{2005}{2005}+\dfrac{1}{2005}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times...\times\dfrac{2006}{2005}=\dfrac{2006}{2}=1003\)
b)\(=\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\times\dfrac{1}{2}=\dfrac{3}{3}\times\dfrac{1}{2}=\dfrac{1}{2}\)
a: \(=68\cdot72\cdot49\cdot\left[324-324\right]=0\)
b: \(=\left(1250-1250\right)\cdot\left(1+...+100\right)=0\)
(1-1/99)x(1-1/100)x(1-1/101)x....x(1-1/2016)
=98/99x99/100x100/101x.....x2015/2016
=98/2016
=7/144
nhớ **** cho mình nha
(1-1/99)*(1-1/100)*...*(1-1/2005)
\(=\frac{98}{99}\cdot\frac{99}{100}\cdot...\cdot\frac{2004}{2005}\)
\(=\frac{98\cdot99\cdot...\cdot2004}{99\cdot100\cdot...\cdot2005}\)
\(=\frac{98}{2005}\)
Kí tên:thangbnshTa 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
\(\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right).....\left(1-\frac{1}{2005}\right)\)
\(=\frac{98}{99}.\frac{99}{100}.....\frac{2004}{2005}=\frac{98}{2005}\)