Question

Simplificando a expressão

E=(1-1/(x+74)^2) (1-1/(x+75)^2) (1-1/(x+76)^2) (1-1/(x+77)^2) (1-1/(x+78)^2)

obtemos uma fração da forma
E= (x + a)(a + b)/(x+74) (x+78)
em que a e b são números inteiros. Qual o valor de a + b?​

Answer 1

Temos a expressão :

$\displaystyle \text E = (1-\frac{1}{(\text x+74)^2}).(1-\frac{1}{(\text x+75)^2}).(1-\frac{1}{(\text x+76)^2}).(1-\frac{1}{(\text x+77)^2}).(1-\frac{1}{(\text x+78)^2}).$

Tirando o MMC de cada parênteses :

$\displaystyle \text E = \frac{((\text x+74)^2-1)}{(\text x+74)^2}.\frac{((\text x+75)^2-1)}{(\text x+75)^2}\frac{((\text x+76)^2-1)}{(\text x+76)^2}\frac{((\text x+77)^2-1)}{(\text x+77)^2}\frac{((\text x+78)^2-1)}{(\text x+78)^2}$

Perceba que em cada parênteses do numerador aparece o produto notável :

$\text x^2-\text y^2 = (\text{x+y})(\text{x-y})$

Por exemplo, no 1º parêntese a = (x+74) e b = 1

$\displaystyle \text E = \frac{(\text x+74+1)(\text x+74-1) }{(\text x+74 )^2} \to \text E = \frac{(\text x+75).(\text x+73)}{(\text x+74 )^2}$

Analogamente, fazendo para todos os termos. Temos :

$\displaystyle \text E = \frac{(\text x+75).(\text x+73)}{(\text x+74 )^2}.\frac{(\text x+76).(\text x+74)}{(\text x+75 )^2}.\frac{(\text x+77).(\text x+75)}{(\text x+76 )^2}.\frac{(\text x+78).(\text x+76)}{(\text x+77)^2}.\frac{(\text x+79).(\text x+77)}{(\text x+78 )^2}$

$\displaystyle \text E = \frac{(\text x+75)^2}{(\text x+75)^2}.\frac{(\text x+76)^2}{(\text x+76)^2}.\frac{(\text x+77)^2}{(\text x+77)^2}.\frac{(\text x+74).(\text x+78).(\text x+79).(\text x+73)}{(\text x+74)^2.(\text x+78)^2}$

$\displaystyle \text E = \frac{(\text x+79).(\text x+73)}{(\text x+74).(\text x+78)}$

Portanto :

$\text a = 79 \ \text e \ \text b = 73$

$\huge\boxed{\text{a+b} = 152 }\checkmark$