If the value of X of fraction x-3 is 0, then x =?

This kind of problem should be transformed into multiplication of numerator and denominator, solving quadratic equation of one variable, and then rounding off some solutions according to denominator not being zero
The answer to this question: x = 0, remember that x cannot be equal to 3

It is known that a root of the quadratic equation AX ^ 2 + 2bx + C = 3 with respect to X is X1 = - 5, and the axis of symmetry of the quadratic function y = ax ^ 2 + BX + C is x = - 5, then what are the vertex coordinates of the image of the quadratic function

It is known that one of the roots of the quadratic equation AX ^ 2 + BX + C = 3 is X1 = - 5
The axis of symmetry of quadratic function y = ax ^ 2 + BX + C is x = - 5
Let the vertex coordinate be (- 5, K)
Then the equation k = 3 is substituted
So the vertex coordinates of this quadratic function image are (- 5,3)

On the quadratic equation with one variable x2 + ax + a = 0 of X, if one root is 3, then the value of a is equal to 0___ ．

Substituting x = 3 into the quadratic equation x2 + ax + a = 0, 9 + 3A + a = 0, the solution is a = - 94

It is known that the quadratic function f (x) = ax square + BX (a, B are constants, and a ≠ 0) satisfies the condition f (x + 1) = f (1-x), and the equation f (x) = x has equal roots
(1) Find the analytic expression of F (x); (2) in the interval [- 1,1], the image of y = f (x) is always above the image of y = 2x + T, try to determine the range of real number T; (3) whether there is real number m, n (m)

1. F (x + 1) = f (1-x), we can know that 4A + 2B = 0
If f (x) = x has equal roots, then (B-1) ^ 2 = 0
So B = 1, a = - 0.5
f（x）=-0.5x^2+x
2. In the interval [- 1,1], if the image of y = f (x) is always above the image of y = 2x + T, then -0.5x ^ 2-x-t does not intersect the X axis on [- 1,1] (note that the symmetry axis of the function image is - 1)
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If the value of X of fraction x-3 is 0, then x =?