Question

GENTE ME AJUDA, URGENTE!!!!!!
SIMLIFIQUE AS FRAÇÕES, ADMITINDO QUE OS DENOMINADORES SEJAM DIFERENTES DE ZERO.

a) (3a-3b) / 12 =
b) (2x+4y) / 2a=
c) (3x-3) / ( 4x -4)
d) ( 3x -3) / ( 3x + 6)
e) ( 5x +10 ) / 5x
f) ( 8x - 8y ) /( 10x - 10y)
g) ( 3a+ 3b ) / ( 6a + 6b)
h) ( 15x²+ 5x) / 5x
i)( 6x - 6y) / ( 3x -3y)
j)(18x - 18 ) / ( 15x - 15)
k)( x² - x) / ( x - 1)
l)( 2x + 2y) / 6

Answer 1

a)  $\frac{3a-3b}{12}=\frac{3(a-b)}{3\times4}=\frac{a-b}{4}$

b)  $\frac{2x+4y}{2a}=\frac{2(x+2y)}{2a}=\frac{x+2y}{a}$

c)  $\frac{3x-3}{4x-4}=\frac{3(x-1)}{4(x-1)}=\frac{3}{4}$

d)  $\frac{3x-3}{3x+6}=\frac{3(x-1)}{3(x+2)}=\frac{x-1}{x+2}$

e)  $\frac{5x+10}{5x}=\frac{5(x+2)}{5x}=\frac{x+2}{x}$

f)  $\frac{8x-8y}{10x-10y}=\frac{2(4x-4y)}{2(5x-5y)}=\frac{4x-4y}{5x-5y}=\frac{4(x-y)}{5(x-y)}=\frac{4}{5}$

g)  $\frac{3a+3b}{6a+6b}=\frac{3a+3b}{2(3a+3b)}=\frac{1}{2}$

h)  $\frac{15x^2+5x}{5x}=\frac{5x(3x+1)}{5x}=\frac{3x+1}{1}=3x+1$

i)  $\frac{6x-6y}{3x-3y}=\frac{2(3x-3y)}{3x-3y}=\frac{2}{1}=2$

j)  $\frac{18x-18}{15x-15}=\frac{3(6x-6)}{3(5x-5)}=\frac{6(x-1)}{5(x-1)}=\frac{6}{5}$

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

l)  $\frac{2x+2y}{6}=\frac{2(x+y)}{2\times3}=\frac{x+y}{3}$