If the square of (a + 1) + the square of (2b-3) + |c-1 | = 0, find the value of the cubic power of 3ABC + a-c

If the square of (a + 1) + the square of (2b-3) + |c-1 | = 0, find the value of the cubic power of 3ABC + a-c

Square of (a + 1) + square of (2b-3) + |c-1 | = 0,
a+1=0,2b-3=0,c-1=0
a=-1,b=1.5,c=1
The third power of 3ABC + a - the third power of C
=The third power of 3 * (- 1) * 1.5 * 1 + (- 1) - the third power of 1
=-4.5-1-1
=-6.5

It is known that the quadratic function f (x) satisfies the condition that f (0) = 0, f (- x + 5) = f (x-3), and the equation f (x) = x has equal roots to find the analytic expression of F (x)

∵ quadratic function f (x) = AX2 + BX, f (- x + 5) = f (x-3) ∵ the axis of symmetry is a straight line x = 1, that is - B / 2A = 1, and f (x) = x has equal roots, that is, ax ^ 2 + bx-x = 0 has equal roots, then (B-1) ^ 2 = 0, so B = 1, then a = - 1 / 2, so f (x) = - 1 / 2x ^ 2 + X ask: why is the axis of symmetry x = 1? Answer: from F (- x + 5) = f (x-3), we can see that the period is 5-3 = 2, the axis of symmetry is half, so it is 1

Let f (x) = 3ax ^ 2 + 2bx + C, if a + B + C = 0, f (0) > 0, f (1) > 0, prove: (I) a > 0 and - 2
Is the axis of symmetry f (- B / 3a), = - (a ^ 2 + C ^ 2-ac) / 3A = [- (A-C) ^ 2 + AC] / 3a

a^2+c^2-ac=a^2+c^2-2ac+ac=(a-c)^2+ac

Let f (x) = 3ax2 + 2bx + C. If a + B + C = 0, f (0) > 0, f (1) > 0, prove: (I) a > 0 and − 2 < ba < 1; (II) the equation f (x) = 0 has two real roots in (0,1)

It is proved that: (I) because f (0) > 0, f (1) > 0, C > 0, 3A + 2B + C > 0. By the condition a + B + C = 0, eliminating B, a > C > 0 is obtained; by the condition a + B + C = 0, eliminating C, a + B < 0, 2A + b > 0 is obtained