Question

encontre as raízes da equação

Answer 1

Resposta: $s=${$-\frac{1}{2}, 5$$}$

Explicação passo-a-passo:

$\frac{x-1}{2} - \frac{x(3-x)}{3}  = x + \frac{1}{3} \\\frac{x-1}{2} - \frac{3x-x^{2}}{3}  = \frac{3x+1}{3} \\\frac{3(x-1) - 2(3x-x^{2})}{6}  = \frac{3x+1}{3} \\\frac{3x-3-6x+2x^{2}}{6}  = \frac{3x+1}{3} \\\frac{-3x-3+2x^{2}}{6}  = \frac{3x+1}{3} \\6(3x+1) = 3(-3x-3+2x^{2})\\18x+6 = -9x-9+6x^{2}\\27x+15-6x^{2}=0\\9x+5-2x^{2}=0\\\\Delta = b^{2}-4ac\\Delta = 9^{2}-4.(-2).5\\Delta = 81+40\\Delta = 121\\\\x=\frac{-b+-\sqrt{Delta}}{2a} \\x=\frac{-9+-\sqrt{121}}{2.(-2)} \\x=\frac{-9+-11}{-4}$$x1=\frac{-9+11}{-4}\\x1=-\frac{1}{2} \\\\x2=\frac{-9-11}{-4}\\x2=5$$Alternativa B.$