The sum of the reciprocals of two consecutive integers is 9/20. Find the integers?
回答 (7)
2017-10-28 4:43 am
1/n + 1/(n+1) = 9/20
(n + 1 + n)/(n * (n + 1)) = 9/20
(2n + 1) * 20 = 9 * (n^2 + n)
9n^2 - 31n - 20 = 0
(n - 4)(9n + 5) = 0
n = -5/9, 4
=> n = 4
(n + 1 + n)/(n * (n + 1)) = 9/20
(2n + 1) * 20 = 9 * (n^2 + n)
9n^2 - 31n - 20 = 0
(n - 4)(9n + 5) = 0
n = -5/9, 4
=> n = 4
2017-10-28 11:59 pm
1/x + 1/(x + 1) = 9/20
(2x + 1)/(x^2 + x) = 9/20
x = 4
The integers are 4 and 5.
(2x + 1)/(x^2 + x) = 9/20
x = 4
The integers are 4 and 5.
2017-10-28 5:28 am
1 / x + 1 / (x + 1) = 9/20
20 (x + 1) + 20 x = (9x)(x + 1)
20x + 20 + 20x = 9x² + 9x
9x² - 31x + 20 = 0
[ 9x + 5 ] [ x - 4 ] = 0
x = - 5/9 , x = 4
Integers are 4 and 5
Check
1/4 + 1/5 = 5/20 + 4/20 = 9/20
20 (x + 1) + 20 x = (9x)(x + 1)
20x + 20 + 20x = 9x² + 9x
9x² - 31x + 20 = 0
[ 9x + 5 ] [ x - 4 ] = 0
x = - 5/9 , x = 4
Integers are 4 and 5
Check
1/4 + 1/5 = 5/20 + 4/20 = 9/20
2017-10-28 5:24 am
(1/x) + (1/(x + 1)) = 9/20
(x + x + 1) / x(x + 1) = 9/20
(2x + 1) / (x^2 + x) = 9/20
20(2x + 1) = 9(x^2 + x)
40x + 20 = 9x^2 + 9x
0 = 9x^2 - 31x - 20
0 = (9x^2 - 36x) + (5x - 20)
0 = 9x(x - 4) + 5(x - 4)
0 = (9x + 5)(x - 4) ---> x = -5/9 or 4.
Disregard -5/9 because that isn't an integer; your two consecutive integers are 4 and 5.
(x + x + 1) / x(x + 1) = 9/20
(2x + 1) / (x^2 + x) = 9/20
20(2x + 1) = 9(x^2 + x)
40x + 20 = 9x^2 + 9x
0 = 9x^2 - 31x - 20
0 = (9x^2 - 36x) + (5x - 20)
0 = 9x(x - 4) + 5(x - 4)
0 = (9x + 5)(x - 4) ---> x = -5/9 or 4.
Disregard -5/9 because that isn't an integer; your two consecutive integers are 4 and 5.
2017-10-28 4:52 am
1/n + 1/(n+1) = 9/20
20(n+1) + 20n = 9n(n+1)
40n + 20 = 9n² + 9n
9n² - 31n - 20 = 0
quadratic formula
n = [31 ± √(31² – 4·9(-20))] / [2·9]
= [31 ± √1681] / 18
= [31 ± 41] /18
= 4, -5/9
-5/9 is an extraneous solution
n = 4
n+1 = 5
20(n+1) + 20n = 9n(n+1)
40n + 20 = 9n² + 9n
9n² - 31n - 20 = 0
quadratic formula
n = [31 ± √(31² – 4·9(-20))] / [2·9]
= [31 ± √1681] / 18
= [31 ± 41] /18
= 4, -5/9
-5/9 is an extraneous solution
n = 4
n+1 = 5
2017-10-28 4:44 am
If one integer is n, the other is of course n + 1. So turning English into maths, we have:
1/n + 1/(n + 1) = 9/20
=> [(n + 1) + n] / n(n + 1) = 9/20
=>20 (2n + 1) = 9n(n + 1)
=> 40n + 20 = 9n² + 9n
=> 9n² - 31n - 20 = 0
=> (n - 4) (9n + 5) = 0
=> n = 4 (because only the integer solution counts).
So the numbers are 4 and 5.
1/n + 1/(n + 1) = 9/20
=> [(n + 1) + n] / n(n + 1) = 9/20
=>20 (2n + 1) = 9n(n + 1)
=> 40n + 20 = 9n² + 9n
=> 9n² - 31n - 20 = 0
=> (n - 4) (9n + 5) = 0
=> n = 4 (because only the integer solution counts).
So the numbers are 4 and 5.
2017-10-28 4:43 am
4 and 5
1/4 + 1/5 = (5+4)/20 = 9/20
1/4 + 1/5 = (5+4)/20 = 9/20
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