Question

resolva as equações

A) x2 - 5x +6 =0
B) x2 - 144=0
C) y2 sobre 2 -y+2 sobre 3 =2y
D) 6x ao quadrado -3x=0

me AJUDA!

Answer 1

a )

x²-5x+ 6 = 0


a = 1 b = -5 c = 6

∆ = b² -4.a.c

∆ = (-5)² -.4.1.6

∆ = 25 -24

∆ = 1


$x = \frac{ - ( - 5)± \sqrt{1} }{2}$


$x1 = \frac{5 + 1}{2} = 3$


$x2 = \frac{5 - 1}{2} = 2$
___________________________________

b)

x² -144 = 0

x² = 144

x = √ 144

x = 12
___________________________


c)

$\frac{ {y}^{2} }{2 - y} + \frac{2}{3} = 2y$

$\frac{ {3y}^{2} }{4 - 2y} = 2y$

3y² = 8y -4y²

3y² +4y² = 8y

7y² = 8y

7y² -8y = 0

y(7y -8) = 0

y1= 0


y2=


7y -8 = 0

7y = 8



$y = \frac{8}{7}$
_______________________

d)
6x² -3x = 0

x (6x-3) = 0

x1 = 0


x2=

6x -3= 0

6x = 3


$x = \frac{3}{6} = \frac{1}{2}$