A = B + 1 (a, B are non-zero natural numbers), the least common multiple of a and B is (?) A. odd number B. even number A = B + 1 (a, B are non-zero natural numbers), the least common multiple of a and B is (?) A. odd number B. even number [please attach the analysis,]

A = B + 1 (a, B are non-zero natural numbers), the least common multiple of a and B is (?) A. odd number B. even number A = B + 1 (a, B are non-zero natural numbers), the least common multiple of a and B is (?) A. odd number B. even number [please attach the analysis,]


Because a = B + 1, they must have an odd number and an even number. The least common multiple of them is a × (B + 1)



1,2,4,8,16,32,64,;...: Law


The rule is that the latter is twice as much as the former
The general formula is: an = 2 ^ (n-1)
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What is 5 out of 12


Multiply it by 60



5 out of 3 × (1 out of 8) + (5 out of 6)


Hello
-5/3×1/8+(-5/6)
=-5/24-5/6
=-25/24
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It is known that the image of a function of degree is parallel to the image of y = - 12x and intersects with the Y axis (0, - 3)


Let the function be y = KX + B, the image of ∵ function is parallel to the image of y = - 12x, ∵ k = - 12, and ∵ the function passes through the point (0, - 3), ∵ 3 = B, ∵ the function is y = - 12x-3



Decompose 60105240 into prime factors and find out the same prime factors


60=2*2*3*5
105=5*5*5
240=2*2*2*2*3*5
The same quality factor is 5



Given A-B = 5, ab = 2, find the value of a + B


(a-b)²=a²-2ab+b²
(a+b)²=a²+2ab+b²=(a-b)²+4ab=25+8=33
a+b=±√33



It is proved that if f (x) has second derivative, and f (0) = f (1) = 0, f (x) / X → 0 (x → 0), then there is at least one point ξ in (0,1), such that f '' (ξ) = 0


If f (x) has a second derivative, then the first derivative of F (x) and f (x) are continuously differentiable
If f (x) / X → 0 (x → 0), then f (x) → 0 (x → 0)
If f (x) is continuous, then (x → 0), f (x) → 0 = f (0) = 0
Then f (x) / X → 0 (x → 0) = [(f (x) - f (0)) / (x-0)] → 0 (x → 0)
That is, f '(0) = 0
Because f (0) = f (1) = 0, according to Rolle's mean value theorem, there is at least one point ξ 1 in (0,1), so that f '(ξ 1) = 0
Because f '(0) = f' (ξ 1) = 0 and the first derivative of F (x) is continuously differentiable
It also satisfies Rolle's mean value theorem
So in (0, ξ 1), i.e. (0,1), there is at least a little ξ, so that f '' (ξ) = 0



Simple calculation question 12x21 × 45 × 102 △ (15 × 4 × 7 × 51)


Take your time



The reciprocal of two fifths is (), and the reciprocal of () is itself


Five out of two, the reciprocal of one is itself