Let the image of function y = f (x) (x ∈ R) be symmetric with respect to line x = 0 and line x = 1, and if x ∈ [0,1], f (x) = X2, then f (− 32)=______ ．

Let the image of function y = f (x) (x ∈ R) be symmetric with respect to line x = 0 and line x = 1, and if x ∈ [0,1], f (x) = X2, then f (− 32)=______ ．

The image of ∵ y = f (x) (x ∈ R) is symmetric with respect to line x = 0 and line x = 1, ∵ f (- x) = f (x), f (1 + x) = f (1-x), ∵ f (− 32) = f (32) = f (1 + 12) = f (1 − 12) = f (12), and when x ∈ [0, 1], f (x) = X2, ∵ f (12) = (12) 2 = 14, that is, f (− 32) = 14

Given the length of the three sides of a positive number a, B, cshi triangle, and the equation (a ^ 2-C ^ 2) + (AB BC) = 0, try to determine the shape of the triangle

Because
(a^2-c^2)+(ab-bc)
=(a-c)(a+c)+b(a-c)
=(a-c)(a+b+c)=0
So a-c = 0 or a + B + C = 0
Only a-c = 0 holds
a=c
So a triangle is an isosceles triangle

It is known that positive numbers a, B and C are the lengths of triangles, and the equation a ^ 2-C ^ 2 + AB BC = 0 holds. Try to determine the shape of the triangle

In △ ABC, D and E are the points on BC and AC respectively, and BD / DC = AE / EC = m / N, then the relation between vector de and vector ab

BD / DC = AE / EC = m / N, then (BD / DC) + 1 = (AE / EC) + 1 = (M / N) + 1, so BC / DC = AC / EC, therefore,
Since △ ABC is similar to △ EDC, de ‖ ab