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x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ... + ( x + 100 ) = 11 000
<=> ( x + x + x + x + x + ... + x ) + ( 1 + 2 + 3 + 4 + ... + 100 ) = 11 000
<=> 101x + \(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)= 11 000 ( vì sao em để 101x thì idol biết mà :33 )
<=> 101x + 5050 = 11 000
<=> 101x = 5950
<=> x = 5950/101
(x+1) + (x+2)+(x+3)+(x+4)+(x+5)=55
=> 5x + 15=55
=>5x=55-15=40
=>x=40/5
Vây x= 8
5.(x - 1) + 6.(x - 2) = 5
5x - 5 + 6x - 12 = 5
11x - 17 = 5
11x = 5 + 17
11x = 22
x = 2
3.(x - 2) + 6.(x - 1) = 6
3x - 6 + 6x - 6 = 6
9x - 12 = 6
9x = 6 + 12
9x = 18
x = 2
5 ( x - 1 ) + 6 ( x - 2 ) = 5
<=> 5x - 5 + 6x - 12 = 5
<=> 11x = 22
<=> x = 2
Vậy x = 2
c=(1/2+1)x(1/3+1)x(1/4+1)x...x(1/99+1)
c=(1-1/2)x(1-1/3)x(1-1/4)x...x(1-1/99)
c1/2x2/3x3/4x...x89/99
c=1/99
a,3x-3+6x-12=5
9x-15=5
9x=20
x=20/9
b,3x-6+6x-6=6
9x-12=6
9x=18
x=2
c,4x+4+5x+10=3x+20
9x+14=3x+20
9x-3x=6
6x=6
x=1
a) 3.(x-1) + 6.(x-2) = 5
3.x - 3 + 6.x - 12 = 5
9.x - 15 = 5
9.x = 5 + 15
9.x = 20
x = \(\frac{20}{9}\)
Vậy x = \(\frac{20}{9}\)
b) 3.(x-2) + 6.(x-1) = 6
3.x - 6 + 6.x - 6 = 6
9.x - 12 = 6
9.x = 6 + 12
9.x = 18
x = 18 : 9
x = 2
Vậy x = 2
c) 4.(x+1) + 5.(x+2) = 3.x +20
4.x +4 +5.x +10 = 3.x +20
9.x + 14 = 3.x +20
9.x - 3.x = 20 - 14
6.x = 6
x = 6 : 6
x = 1
Vậy x = 1
Ta có
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times....\times\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{9}{10}\)
\(=\frac{1}{10}\)
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)\)
= \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x\frac{5}{6}\)
=\(\frac{1x2x3x4x5}{2x3x4x5x6}\)
Loại 2x3x4x5 vì cả 2 vế cùng có
=\(\frac{1}{6}\)
\(\left(1-\frac{1}{2}\right)\) x \(\left(1-\frac{1}{3}\right)\)x \(\left(1-\frac{1}{4}\right)\)x \(\left(1-\frac{1}{5}\right)\)x \(\left(1-\frac{1}{6}\right)\)
\(=\)\(\frac{1}{2}\) x \(\frac{2}{3}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\)x \(\frac{5}{6}\)
\(=\)\(\frac{1}{6}\)
c) (x+1) + (x+2) + ... + (x+5) = 90
=> 5x + ( 1 + 2 + ... + 5 ) = 90
5x + 15 = 90
5x = 90 - 15
5x = 75
x = 75 : 5
x = 15
d) (x+1) + (x+2) + .... + (x+100) = 20150
=> 100x + ( 1+2+...+100 ) = 20150
100x + 5050 = 20150
100x = 20150 - 5050
100x = 15100
x = 15100 : 100
x = 151
Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90
<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90
<=> 5x + 15 = 90
=> 5x = 75
=> x = 15