Question

Arătați ca E(x)=$\frac{2(3-x)}{x+3}$
E(x)= $\frac{3-x}{3+x} + \frac{3+x}{3-x} + \frac{12x}{ x^{2}-9 }$

Answer 1

E(x) = 2(3-x)/x+3
E(x) = 2*3-2x/x+3
E(x) = 6-2x/x+3

E(x) = 3-x/3+x+3+x/3-x+12x/x²-9
E(x) = 3-x/3+x-3+x/-(x-3)+12x/(x-3)(x+3)
E(x) = 3-x/3+x-3+x/x-3+12x/(x-3)(x+3)
E(x) = (x-3)(3-x)-(3+x²)+12x/(x-3)(3+x)
E(x)= (x-3)*(-(x-3))-(9+6x+x²)+12x/(x-3)(3+x)
E(x) = -(x-3)²-9-6x-x²+12x/(x-3)(3+x)
E(x) = -(x²-6x+9)-9-6x-x²+12x/(x-3)(3+x)
E(x) = -x²+6x-9-9-6x-x²+12x/(x-3)(3+x)
E(x) = -2x²-18+12x/(x-3)(3+x)
E(x) =-(2x²-12x+18)/(x-3)(3+x) = -2(x²-6x+9)/(x-3)(3+x)
E(x) = -2(x-3)²/(x-3)(3+x) = -2(x-3)/3+x = -2x+6/3+x