(x+1)+(x+2)+(x+3)+.........+(x+99)=5000
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Ta có:
( x + 1 ) + ( x + 3 ) + ( x + 5 ) + .... + ( x + 99 ) = 5 000
=> x + 1 + x + 3 + x + 5 +.....+ x + 99 = 5 000
=> ( x + x + x +....+ x ) + ( 1 + 3 + 5 +.....+ 99 ) = 5 000
=> 50x + 2 500 = 5 000
=> x = ( 5 000 - 2 500 ) : 50
=> x = 50
50x + 2500 = 5000
50x = 5000 - 2500
50x = 2500
x = 2500 : 50
x = 50
1 a) (x+ 1) + (x + 2 ) + (x + 3) + ... + (x + 100) = 205550 (100 cặp)
=> (x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 205 550
100 số hạng x 100 số hạng
=> 100.x + 100 . 101 : 2 = 205 550
=> 100.x + 5050 = 205 550
=> 100 . x = 205 550 - 5050
=> 100 . x = 200500
=> x = 200500 : 100
=> x = 2005
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5000 + 100 = 5100
7400 - 400 = 7000
2000 3 + 600 = 6600
8000 : 2 + 2000 = 6000
6000 - (5000 - 1000) = 2000
6000 - 5000 + 1000 = 2000
7000 - 3000 x 2 = 1000
(7000 - 3000) x 2 = 8000
5000 + 100 = 5100
7400 - 400 = 7000
2000 3 + 600 = 6600
8000 : 2 + 2000 = 6000
6000 - (5000 - 1000) = 2000
6000 - 5000 + 1000 = 2000
7000 - 3000 x 2 = 1000
(7000 - 3000) x 2 = 8000
= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10
= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10
=1-1/10=9/10
đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10
\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2\times\frac{99}{100}\)
\(=\frac{99}{50}\)
(x+1)+(x+2)+(x+3)+...+(x+99)=5000
=> x.[(99-1):1+1]+(1+2+3+...+99)=5000
=>99x+{(99+1).[(99-1):1+1]:2}=5000
99x+4950=5000
99x=5000-4950=50
x=50/99
Vậy x=50/99
(x+1)+(x+2)+(x+3)+.....+(x+99)=5000
99.x+(1+2+3+.....+99)=5000
99.x+{(99+1).[(99-1).1+1]:2}=5000
99.x+5000=5000
99.x=5000-5000
99.x=0
x=0:99
x=0