• Matéria: Matemática
  • Autor: SUPREME75
  • Perguntado 3 anos atrás


 \frac{2x - 3}{x + 6}  =  \frac{3x - 1}{x - 2}
( "X" diferente de "6" e "X" diferente de "2" ).​

Anexos:

Respostas

respondido por: andferg
1

Resposta:

x_{1} = -12-2\sqrt{39} e x_{2} = -12+2\sqrt{39}

Explicação passo a passo:

\begin{aligned}\frac{2x-3}{x+6} = \frac{3x-1}{x-2} & \iff (2x-3)(x-2) = (3x-1)(x+6)\\& \iff x^{2}+24x-12 = 0 \iff x_{1,2} = \frac{-24\pm\sqrt{24^{2}-4(1)(-12)}}{2}\\& \iff x_{1,2} = \frac{-24\pm \sqrt{624}}{2} \iff x_{1,2} = \frac{-24\pm 4\sqrt{29}}{2} \\& \iff x_{1,2} = \frac{-12\pm2\sqrt{39}}{2} \iff \begin{cases} x_{1} = \frac{-12 - 2\sqrt{39}}{2} \\ x_{2} = \frac{-12 + 2\sqrt{39}}{2} \end{cases}\end{aligned}

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