12 divided by (1 / 4-1 / 3) - 2.5 divided by (- 1 / 4)

12 divided by (1 / 4-1 / 3) - 2.5 divided by (- 1 / 4)

solution
simple form
=12÷(3/12-4/12)-2.5÷(-0.25)
=12÷(-1/12)+10
=12×(-12)+10
=-144+10
=-134

If x > 0, the minimum value of F (x) = 12x + 3x & nbsp; is ()
A. 12B. -12C. 6D. -6

Because x > 0, f (x) = 12x + 3x ≥ 212x · 3x = 12. Therefore, the minimum value of F (x) = 12x + 3x & nbsp; is 12

When x > 1, find the minimum value of F (x) = (x ^ 2 + 3x + 9) / (x-1)

Let X-1 = t (T > 0)
Then x = t + 1
y=[(t+1)²+3(t+1)+9]/t
=(t²+5t+13)/t
=t+13/t+5
≥2√13 +5
If and only if t = √ 13, i.e. x = 1 + √ 13, the equal sign holds
Therefore, the minimum value of F (x) = (x ^ 2 + 3x + 9) / (x-1) is 2 √ 13 + 5

If the function & nbsp; f (x) = 13x3-x has the minimum value in the interval (1-A, 10-a2), then the value range of real number a is___ ．

The function f (x) = 13x3-x, so f ′ (x) = x2-1; F (x) = 13x3-x has the minimum value in the interval (1-A, 10-a2), so the function f (x) first decreases and then increases in the interval (1-A, 10-a2), that is, f ′ (x) is less than 0 and then greater than 0, so combining with the properties of quadratic function, we can get: 1-A < 1 < 10-a2, and f (1) = - 23 = f (- 2), then 1-A ≥ - 2 solution: 0 < a ≤ 1