Given a ＞ B ＞ 0, find the minimum value of a 2 + 16b (a − b)

Given a ＞ B ＞ 0, find the minimum value of a 2 + 16b (a − b)

∵ B (a-b) ≤ (B + a − B2) 2 = A24, ∵ A2 + 16b (a − b) ≥ A2 + 64a2 ≥ 16. If and only if B = a − Ba2 = 8, i.e. a = 22b = 2, the equal sign is taken

When AB is the value, the polynomial a ^ 2 + B ^ 2-8a + 10B + 41 has the minimum value

a²+b²-8a+10b+41
=a²-8a+16+b²+10b+25
=(a-4)²+(b+5)
therefore
When a = 4, B = - 5, there is a minimum, and the minimum is 0

For the square of algebraic formula 1 / 2A - A + B = 0, its value is 1, then when a = 2, its value is ()

If the equation is 1 / 2 * 1 square - 1 + B = 0, the solution is b = 1 / 2, and substituting it into the original formula is 1 / 2A square - A + 1 / 2 = 0, then substituting a = 2 into the original formula is - 1 / 2

Let f (x) be an odd function on R, but when x is greater than 0, f (x) = x (the root of 1 + X to the third power) to find the expression of F (x) on R

Let x0
F(-x) = (-x)(1+(-x)^3/2)
Because it is an odd function
F(-x) = -F(x) = (-x)(1+(-x)^3/2)
F(x) = x(1+(-x)^3/2)
When x = 0,
∵F(x) = -F(x),F(0) = 0
Overall:
F(x) =
x(1+(-x)^3/2) x0
0 x=0