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Ta có:(x + 1)+(x+3)+...+(x+9)=75
=>5x + (1+3+...+9)=75
=>5x + 25=75
=>5x =50
=>x = 10
a. (x+1)+(x+3)+(x+5)+(x+7)+(x+9)=75
=>(x+x+x+x+x)+(1+3+5+7+9)=75
=>Xx5+25=75
=>Xx5=75-25
=>Xx5=50
=>x=50:5
=>x=10
b. sai đề
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) (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
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
(x+1) + (x+2)+(x+3)+(x+4)+(x+5)=55
=> 5x + 15=55
=>5x=55-15=40
=>x=40/5
Vây x= 8
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)\)
= \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x\frac{5}{6}\)
=\(\frac{1x2x3x4x5}{2x3x4x5x6}\)
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 \(\frac{5}{6}\)
\(=\)\(\frac{1}{6}\)