Question

Dadas as funções f(x)= x+1/x-2, g(x)= 2x-1/x+2 e h(x)=5, resolva a equação f(x) +g(x)=h(x)

Answer 1

Olá.

Basta substituirmos os valores de F(x), G(x) e H(x).
Vamos aos cálculos:
$\mathsf{F(x)+G(x)=H(x)}\\\\ \mathsf{\dfrac{x+1}{x-2}+\dfrac{2x-1}{x+2}=5}\\\\\\ \mathsf{\dfrac{(x+1)(x+2)+(2x-1)(x-2)}{(x-2)(x+2)}=5}\\\\\\ \mathsf{\dfrac{(x^2+2x+x+2)+(2x^2-4x-x+2)}{(x^2+2x-2x-4)}-5=0}\\\\\\ \mathsf{\dfrac{x^2+2x+x+2+2x^2-4x-x+2-5(x^2-4)}{(x^2-4)}=0}\\\\\\ \mathsf{\dfrac{x^2+2x^2+2x+x-4x-x+2+2-(5x^2-20)}{(x^2-4)}=0}\\\\\\ \mathsf{3x^2-2x+4-(5x^2-20)=0\cdot(x^2-4)}\\\\\\ \mathsf{3x^2-2x+4-5x^2+20=0}\\\\\\ \mathsf{3x^2-5x^2-2x+4+20=0}\\\\\\ \mathsf{-2x^2-2x+24=0}$

Chegamos em uma equação de 2° grau. Para resolvermos e encontrar as duas raízes possíveis, vamos usar Bhaskara. Vamos aos cálculos.
Usaremos a forma ax² + bx + c = 0 para encontrar os coeficientes.
$\mathsf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\\\\ \mathsf{x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot(-2)\cdot24}}{2(-2)}}\\\\\\ \mathsf{x=\dfrac{+2\pm\sqrt{4-4\cdot(-48)}}{-4}}\\\\\\ \mathsf{x=\dfrac{+2\pm\sqrt{4+192}}{-4}}\\\\\\ \mathsf{x=\dfrac{+2\pm\sqrt{196}}{-4}}\\\\\\ \mathsf{x=\dfrac{+2\pm14}{-4}}$

Vamos agora encontrar as duas raízes possíveis.
$\mathsf{x'=\dfrac{+2+14}{-4}}\\\\\\ \mathsf{x'=\dfrac{16}{-4}}\\\\ \boxed{\mathsf{x'=-4}}\\\\\\ \\\mathsf{x''=\dfrac{+2-14}{-4}}\\\\\\ \\\mathsf{x''=\dfrac{-12}{-4}}\\\\ \boxed{\mathsf{x''=3}}\\\\\\\\\boxed{\mathsf{S=\{x\in\mathbb{R}~|~-4,3\}}}$

Qualquer dúvida, deixe nos comentários.
Bons estudos.