有關數學最小值問題
回答 (2)
2007-03-16 7:54 am
✔ 最佳答案
Notice that since a and b are positive, so we could geta = √21 * √b = √3 * √7 * √b
Since √3 and √7 are irreducible, so to take a being a positive integer, we must have
b = 3 * 7 = 21
i.e. The least value of b is 21
2007-03-16 5:00 am
a^2=21b, then b=a^2/21
since b and a must be a whole +ve no, then
a^2 must be at least 21 in order to make a^2/21 a whole no.
So a must be 21, and b=1.
The smallest value for b is 1
since b and a must be a whole +ve no, then
a^2 must be at least 21 in order to make a^2/21 a whole no.
So a must be 21, and b=1.
The smallest value for b is 1
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