Question

f(x)=$\frac{2x-1}{5}$

Aduceti la cea mai simpla forma expresia:

E(x)=$\frac{5|f(x)|^2-5f(x)}{x^2-4x+3}$

Answer 1

Numitorul se scrie astfel:
x^2-x-3x+3=x(x-1)-3(x-1)=(x-1)(x-3)

Numaratorul se scrie astfel:
5(2x-1)^2:25-5(2x-2):5=(4x^2-4x+1)/5-(2x-1)=(4x^2-4x+1-10x+5)/5=
=(4x^2-14x+6)/5=(4x^2-12x-2x+6)/5=[4x(x-3)-2(x-3)]/5=(x-3)(4x-2)/5=
=2(2x-1)(x-3)/5

Observam ca se simplifica  prin x-3 si obtinem:
E(x)=2(2x-1)/[5(x-1)]

Answer 2

E(x)= 5* | (2x-1)/5 |² -5 * (2x-1)/5  supra x² -x -3x+3

= 5* (2x-1)²/25 - 5* (2x-1)/5   supra  x(x-1) -3(x-1)=

= (2x-1)²/5 - 5*(2x-1)/5 supra (x-1)(x-3)=        aducem la acelasi numitor la numarator

= [(2x-1)²-5(2x-1)]/5  supra (x-1)(x-3)=

=(2x-1)[(2x-1)-5]/5  supra (x-1)(x-3)=

= (2x-1)(2x-6)/5  * 1/ (x-1)(x-3)=

=2(2x-1)(x-3)/5 * 1/(x-1)(x-3)=

= 2(2x-1)/5 * 1/(x-1)=

=2(2x-1)/5(x-1)