Question

efectuati respectand ordinea efectuarii operatiilor. (x-3)^2+(3x+4)^2-(2x+3)(2x+3)-6x(x+3)

Answer 1

=x²-6x+9+9x²+24x+16-(2x+3)(x+3)-6x(x-3)= x²-6x+9+9x²+24x+16-(2x+3)²-(6x²-9x)= x²-6x+9+9x²+24x+16-4x²-12x-9-6x²+9x=(cele cu x² sevrefuc, caci de 10x² si -10x²)=
15x+16

Answer 2

[tex]=x^2-6x+9+9x^2+24x+16-(2x+3)(2x+3)-6x(x-3)\\\\ =x^2-6x+9+9x^2+24x+16-4x^2-12x-9-6x^2+9x\\\\ =x^2+9x^2-4x^2-6x^2-6x+24x-12x+9x+9+16-9\\\\ \bold{\underline{=15x+16}}[/tex]