Question

simplifique as frações algébricas,considerando os denominadores diferentes de zero

Answer 1

Explicação passo-a-passo:

a)

$\sf \dfrac{5a-10}{5a}=\dfrac{5\cdot(a-2)}{5a}=\red{\dfrac{a-2}{a}}$

b)

$\sf \dfrac{3x-3y}{6x-6y}=\dfrac{3\cdot(x-y)}{6\cdot(x-y)}=\dfrac{3}{6}=\red{\dfrac{1}{2}}$

c)

$\sf \dfrac{a^2-9}{5\cdot(a+3)}=\dfrac{(a-3)\cdot(a+3)}{5\cdot(a+3)}=\red{\dfrac{a-3}{5}}$

d)

$\sf \dfrac{3a^2-b^2}{3a^2-6ab+3b^2}$

$\sf =\dfrac{3\cdot(a^2-b^2)}{3\cdot(a^2-2ab+b^2)}$

$\sf =\dfrac{3\cdot(a-b)\cdot(a+b)}{3\cdot(a-b)\cdot(a-b)}$

$\sf =\red{\dfrac{a+b}{a-b}}$

e)

$\sf \dfrac{ma-mb}{a\cdot(a-b)-b\cdot(a-b)}$

$\sf =\dfrac{m\cdot(a-b)}{(a-b)\cdot(a-b)}$

$\sf =\red{\dfrac{m}{a-b}}$

Answer 2

Caso tenha problemas para visualizar a resposta experimente abrir pelo navegador.

a)

$\tt \dfrac{5a-10}{5a}=\dfrac{\diagup\!\!\!5\cdot(a-2)}{\diagup\!\!\!5a}={\dfrac{a-2}{a}}$

b)

$\sf \dfrac{3x-3y}{6x-6y}=\dfrac{3\cdot\diagdown\!\!\!\!\!(x\diagdown\!\!\!-\diagdown\!\!\!y)}{6\cdot\diagdown\!\!\!\!(x\diagdown\!\!\!\!-\diagdown\!\!\!y)}=\dfrac{\diagup\!\!\!3^1}{\diagup\!\!\!6_2}=\dfrac{1}{2}\checkmark}$

c)

$\tt \dfrac{a^2-9}{5\cdot(a+3)}=\dfrac{\diagdown\!\!\!\!\!(a\diagdown\!\!\!\++\diagdown\!\!\!3)\cdot(a-3)}{5\cdot\diagdown\!\!\!\!\!(a\diagdown\!\!\++\diagdown\!\!\!3)}={\dfrac{a-3}{5}\checkmark}$

d)

$\sf \dfrac{3a^2-b^2}{3a^2-6ab+3b^2}$

$\sf =\dfrac{3\cdot(a^2-b^2)}{3\cdot(a^2-2ab+b^2)}$

$\sf =\dfrac{\diagup\!\!\!3\cdot\diagdown\!\!\!\!\!(a\diagup\!\!\!-b)\cdot(a+b)}{\diagup\!\!\!3\cdot\diagdown\!\!\!\!\!(a\diagup\!\!\!\!-b)\cdot(a-b)}\sf ={\dfrac{a+b}{a-b}\checkmark}$

e)

$\tt \dfrac{ma-mb}{a\cdot(a-b)-b\cdot(a-b)}\tt =\dfrac{m\cdot\diagup\!\!\!\!\!\!(a\diagdown\!\!\!\!-b)}{(a-b)\cdot\diagup\!\!\!\!\!\!(a\diagdown\!\!\!\!\!-b)}\tt =\red{\dfrac{m}{a-b}\checkmark}$