Question

AJUTATIMA VAROG DAU 40 PUNTE!!!
$\frac{1}{x+1}+ \frac{2}{x+2}+ \frac{3}{x+3}=3$

Answer 1

(x+2)(x+3)+2(x+1)(x+3)+3(x+1)(x+2)/(x+1)(x+2)(x+3)=3(x+1)(x+2)(x+3)/(x+1)(x+2)(x+3)
<=> (x+2)(x+3)+2(x+1)(x+3)+3(x+1)(x+2)=3(x+1)(x+2)(x+3)<=>
<=>x^2+3x+2x+6+2(x^2+3x+x+3)+3(x^2+2x+x+2)=3(x^2+2x+x+2)(x+3)<=>
<=>x^2+3x+2x+6+2x^2+6x+2x+6+3x^2+6x+3x+6=3(x^3+3x^2+2x^2+6x+x^2+3x+2x+6)<=>
<=>x^2+3x+2x+6+2x^2+6x+2x+6+3x^2+6x+3x+6=3x^3+9x^2+6x^2+18x+3x^2+9x+6x+18<=>
<=>6x^2+22x+18=3x^3+18x^2+33x+18<=>
6x^2+22x+18-3x^3-18x^2-33x-18=0<=>
-3x^3-12x^2-11x=0<=>
<=>-x(3x^2+12x+11)=0<=>
x=0
3x^2+12x+11=0
d=b^2-4ac
d=144-132
d=12
rad d=2rad din3
x=(-b+-rad d)/2a
x=(12+2rad din3)/6
x=(6+rad din 3)/3
x=(12-2rad din 3)/6
x=(6-rad din3)/3