Question

36.   Sendo A = ab+ac+6b+6c/3ab+3ac
e B = 3a+3b/a²+ab, calcule:

      
  

 

        a)    A+B                                                              
b)    A-B                                                                               

Answer 1

Olá, Lalá !

$A=\dfrac{ab+ac+6b+6c}{3ab+3ac}$

$B=\dfrac{3a+3b}{a^2+ab}$

Observe que, $ab+ac=a(b+c)$ e $6b+6c=6(b+c)$.

Assim, $ab+ac+6b+6c=a(b+c)+6(b+c)=(a+6)(b+c)$

Além disso, $3ab+3ac=3a(b+c)$. Com isso:

$A=\dfrac{ab+ac+6b+6c}{3ab+3ac}=\dfrac{(a+6)(b+c)}{3a(b+c)}=\dfrac{a+6}{3a}$.

Analogamente, temos:

$3a+3b=3(a+b)$ e $a^2+ab=a(a+b)$.

Deste modo, $B=\dfrac{3a+3b}{a^2+ab}=\dfrac{3(a+b)}{a(a+b)}=\dfrac{3}{a}$.

Logo:

$A+B=\dfrac{a+6}{3a}+\dfrac{3}{a}=\dfrac{a+6+9}{3a}=\dfrac{a+15}{3a}$

$A-B=\dfrac{a+6}{3a}-\dfrac{3}{a}=\dfrac{a+6-9}{3a}=\dfrac{a-3}{3a}$

Espero ter ajudado, até mais ^^