Question

alguém pode me responder a essa questão

Answer 1

Resposta:

a) $\frac{11x-12}{\left(x-3\right)^2}$

b) $\frac{4y-7}{\left(y+3\right)\left(y-3\right)}$

c) $2$

Explicação passo a passo:

a) $\frac{7x}{x^2-6x+9}+\frac{4}{x-3}$

$=\frac{7x}{\left(x-3\right)^2}+\frac{4}{x-3}$

$=\frac{7x}{\left(x-3\right)^2}+\frac{4\left(x-3\right)}{\left(x-3\right)^2}$

$=\frac{7x+4\left(x-3\right)}{\left(x-3\right)^2}$

$=\frac{11x-12}{\left(x-3\right)^2}$

b) $\frac{5}{y^2-9}+\frac{4}{y+3}$

$=\frac{5}{\left(y+3\right)\left(y-3\right)}+\frac{4}{y+3}$

$=\frac{5}{\left(y+3\right)\left(y-3\right)}+\frac{4\left(y-3\right)}{\left(y+3\right)\left(y-3\right)}$

$=\frac{5+4\left(y-3\right)}{\left(y+3\right)\left(y-3\right)}$

$=\frac{4y-7}{\left(y+3\right)\left(y-3\right)}$

c) $\frac{x+y}{3a}\cdot \frac{6ax-6ay}{x^2-y^2}$

$=\frac{x+y}{3a}\cdot \frac{6a}{x+y}$

$=\frac{\left(x+y\right)\cdot \:6a}{3a\left(x+y\right)}$

$=\frac{6a}{3a}$

$=\frac{6}{3}$

$=2$