Question

Resolva as equações em R.
k) (2x - 3)+ 12x = 9
l) 3x²/4 -5x = 0
( nessa precisa de m.m.c antes de resolver)

Answer 1

Oie, Td Bom?!

k)

$(2x - 3) + 12x = 9$

$2x - 3 + 12x = 9$

$14x - 3 = 9$

$14x = 9 + 3$

$14x = 12$

$x = \frac{12}{14}$

$x = \frac{12 \div 2}{14 \div 2}$

$x = \frac{6}{7}$

$S = \left \{ \frac{6}{7} \right \}$

l)

$\frac{3x {}^{2} }{4} - 5x = 0$

$4( \frac{3x {}^{2} }{4} - 5x) = 4 \: . \: 0$

$4 \: . \: \frac{3x {}^{2} }{4} - 4 \: . \: 5x = 0$

$3x {}^{2} - 20x = 0$

$x \: . \: (3x - 20) = 0$

• $x = 0$

• $3x - 20 = 0⇒3x = 20⇒x = \frac{20}{3}$

$S = \left \{0 \: , \: \frac{20}{3} \right \}$

Att. Makaveli1996

Answer 2

Vamos lá:

(2x - 3) + 12 = 9

2x - 3 + 12x = 9

2x + 12x = 9 + 3

14x = 12

x = 12/14 = 6/7

3x²/4 - 5x  = 0   >>> Mmc(1 , 4) = 4

3x² - 20x = 0

x(3x - 20)= 0

x₁ = 0 ; x₂ = 20/3

Espero ter ajudado !!!