Question

preciso de ajuda, tem que ser feito em formula de Bhaskara.​

Answer 1

Resposta:

S={1,5}

Explicação passo-a-passo:

mmc(4,6,12)

4, 6, 12 | 2

2, 3, 6 | 2

1, 3, 3 | 3

1, 1, 1 | 1

MMC = 2 . 2 . 3 = 12

[3x(x+1)-(x-5)]/12=[10(2x-1)]/12

3x²+3x-x+5=20x-10

3x²-18x+15=0  (÷3)

x²-6x+5=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-6x+5=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-6~e~c=5\\\\\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(1)(5)=36-(20)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-6)-\sqrt{16}}{2(1)}=\frac{6-4}{2}=\frac{2}{2}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-6)+\sqrt{16}}{2(1)}=\frac{6+4}{2}=\frac{10}{2}=5\\\\S=\{1,~5\}$