Question

Alguém pode me dizer como simplificar isso?​

Answer 1

E)
= 1/x² - 1 : (x-1)
= (1 - x²)/x² : (x-1)
= (1 - x²)/x² . 1/(x-1)
= (1-x).(1+x)/x² . 1/(-1).(-x+1)
= - (1-x)/(1-x) . (1+x)/x²
= - (1+x)/x²

G)
= (1/x - 1/5) : (x - 5)
= (5 - x)/5x . 1/(x-5)
= 1.(5-x)/5x . (-1)/(-X+5)
= - (5-x)/5x . 1/(5-x)
= -1/5x

I)
= (1/x² - 1/p²) : (x-p)
= (p² - x²)/x².p² . 1/(x-p)
= (p+x).(p-x)/x².p² . 1/-1(-x+p)
= - (P-x)/(p-x) . (P+x)/x².p²
= - (p+x)/x².p²

Answer 2

Resposta:

$e)\frac{\frac{1}{x^2}-1 }{x-1} =\frac{\frac{1-x^2}{x^2} }{x-1}=\frac{1-x^2}{x^2(x-1)} =-\frac{(1+x)(1-x)}{x^2(1-x)} =-\frac{1+x}{x^2}$

$g)\frac{\frac{1}{x}-\frac{1}{5} }{x-5}=\frac{\frac{5-x}{5x} }{x-5} =\frac{5-x}{5x}*\frac{1}{x-5} =-\frac{x-5}{5x} *\frac{1}{x-5}=-\frac{1}{5x}$

$i)\frac{\frac{1}{x^2}-\frac{1}{p^2} }{x-p}=\frac{\frac{p^2-x^2}{x^2p^2} }{x-p}=\frac{\frac{(p+x)(p-x)}{x^2p^2} }{x-p}=\frac{(p+x)(p-x)}{x^2p^2} *\frac{1}{x-p} =-\frac{(p+x)(x-p)}{x^2p^2} *\frac{1}{x-p} =-\frac{p+x}{x^2p^2}$