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gabelawson6996
05.10.2019 •
Mathematics
Find three consecutive even integers. such that 7 times of first integer is 14 more than the sum of second and third integers.
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Ответ:
Given Condition is
7(2n)=(2n+2+2n+4)+14
14n=4n+6+14
14n-4n=20
10n=20
n=20/10
n=2
So the Numbers are 2x2, 2x2+2, 2x2+4
Answer is 4,6,8
Ответ:
(2x+3)(3x+2)
First: 2x•3x=6x^2 Inner: 3x•3=9x
Outer: 2x•2=4x Last:3•2=6
6x^2+9x+4x+6= 6x^2+13x+6