Question

resolva$\frac{10x}{x-1} - \frac{2x-4}{x+1} = - \frac{4}{ x²-1} :$

Answer 1

Tendo em mente a seguinte propriedade

$\large\fbox{ \dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd} }$

Lembrando também do seguinte produto notável:

$\large\fbox{ a^2-b^2=(a+b)\cdot(a-b) }$

Portanto

$\dfrac{10x}{x-1}-\dfrac{2x-4}{x+1}=-\dfrac{4}{x^2-1}\\\\\\\dfrac{[10x\cdot(x+1)]-[(x-1)\cdot(2x-4)]}{(x-1)\cdot(x+1)}=-\dfrac{4}{x^2-1}\\\\\\\dfrac{[10x^2+10x]-[2x^2-4x-2x+4]}{x^2-1}=-\dfrac{4}{x^2-1}\\\\\\\dfrac{10x^2+10x-2x^2+4x+2x-4}{x^2-1}=-\dfrac{4}{x^2-1}\\\\\\\dfrac{8x^2+16x-4}{x^2-1}=-\dfrac{4}{x^2-1}\\\\\\\dfrac{8x^2+16x-4}{x^2-1}+\dfrac{4}{x^2-1}=0\\\\\\\dfrac{8x^2+16x-4+4}{x^2-1}=0\\\\\\\dfrac{8x^2+16x}{x^2-1}=0\\\\\\\underbrace{8x^2+16x=0}_{equa.~quadr\'atica}$

Resolvendo a equação quadrática pela fórmula de Bhaskara

$\dfrac{-b\pm \sqrt{\Delta}}{2a}~~~~onde~~\Delta=b^2-4ac$

Primeiro vamos resolver o discriminante Δ

$\Delta=16^2-(4\cdot8\cdot0)\\\\\Delta=256-0\\\\\Delta=256$

Substituindo Δ na fórmula

$\dfrac{-16\pm \sqrt{256}}{2\cdot8}\\\\\\\dfrac{-16\pm16}{16}\\\\\\x_1=\dfrac{-16+16}{16}=0\\\\\\x_2=\dfrac{-16-16}{16}=\dfrac{-32}{16}=-2\\\\\\\\Solu\c{c}\~ao:\left\{\begin{matrix}x=-2\\\\ou\\\\{\hspace{-5}x=0}\end{matrix}\right.$


$S=\begin{Bmatrix}-2,0\end{Bmatrix}$