Question

Cum se face exercitiul e) ? DAU COROANA!

Answer 1

-3 < x < 3 <=> x²-9 < 0 <=> |x²-9|=9-x² ; |-x²|=x² ;

9-x²+x²+6x+6=15+6x ;

16-x² > 0 <=> |16-x²|=16-x² ; |x+5|·|x+4|=(x+5)·(x+4)=x²+4x+5x+20=x²+9x+20 pentru 0 <= x < 3 si |x+5|·|x+4|=(5-x)·(4-x)=20-5x-4x-x²=20-9x-x² pentru -3 < x < 0 .

Asadar, trebuie sa luam in calcul doua variante

i.  -3 < x <0 => 9+6x+6 /16-x²+20-9x-x²+3x =15+6x /36-2x²-6x ;

ii.  0 <= x < 3 => 9+6x+6 /16-x²+x²+9x+20+3x =15+6x /36+12x =5+2x /12+4x .