Question

$-\frac{5(a+4)}{a^2-2a-8}-\frac{a}{a^2+5a+6}$
Ajutati-maaa va rooggg frumos!!!Cineee ma ajuta,imi trebuie pe miine,urgent
Ajutati-maa va roog

Answer 1

Faci urmatoarele descompuneri!
a^2 + 5a + 6 = (a + 2)(a + 3);
a^2 - 2a - 8 = (a + 2)(a - 4);
Numitorul comun este (a+2)(a+3)(a-4);
Aplifici prima fractie cu (a+3) si a doua fractie cu (a-4);
Obtii la numaratorul cumulat al celor doua fractii asa: -5(a+4)(a+3) - a(a-4);
adica, -5(a^2 + 4a + 3a + 12) - a^2 + 4a = -5a^2 - 35a -  60 - a^2 + 4a = 
-6a^2 - 31a - 60; acesta are delta negativ, asadar nu il putem descompune in  R;
La final obtii:( -6a^2 - 31a - 60)/[(a+2)(a+3)(a-4)];
Bafta!

Answer 2

   
[tex]\displaystyle \\ -\frac{5(a+4)}{a^2-2a-8}-\frac{a}{a^2+5a+6}=? \\ \\ \texttt{Descompunem numitorii in factori primi.} \\ \\ a^2-2a-8 = \underbrace{a^2+2a}-\underbrace{4a-8}=a(a+2)-4(a+2) =(a+2)(a-4) \\ \\ a^2+5a+6 = \underbrace{a^2+2a}+\underbrace{3a+6}=a(a+2)+3(a+2)= (a+2)(a+3) \\ \\ -\frac{5(a+4)}{a^2-2a-8}-\frac{a}{a^2+5a+6}= \\ \\ =-\frac{5(a+4)}{(a+2)(a-4) }-\frac{a}{(a+2)(a+3)}= \\ \\ =-\frac{5(a+4)(a+3)}{(a+2)(a-4)(a+3) }-\frac{a(a-4)}{(a+2)(a+3)(a-4)} = [/tex]


[tex]\displaystyle \\ =\frac{-5(a+4)(a+3) - a(a-4)}{(a+2)(a-4)(a+3)} = \\ \\ =\frac{-5(a^2+7a +12) - (a^2-4a)}{(a+2)(a-4)(a+3)} = \\ \\ =\frac{-5a^2-35a -60 - a^2+4a}{(a+2)(a-4)(a+3)} = \\ \\ =\frac{-6a^2-31a -60}{(a+2)(a-4)(a+3)} = \boxed{-\frac{6a^2+31a +60}{(a+2)(a-4)(a+3)} } [/tex]