Question

(UFPA)
$\sf \dfrac{2x + 1}{x - 3} - \dfrac{2}{x^2 - 9} = 1$

Answer 1

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$\sf{\dfrac{2x+1}{x-3}-\dfrac{2}{x^2-9}=1~~com~x\ne3~e~x\ne-3.}\\\sf{\dfrac{2x+1}{x-3}-\dfrac{2}{(x-3)\cdot(x+3)}=1\times(x-3)\times(x+3)}\\\sf{(2x+1)(x+3)-2=x^2-9}\\\sf{2x^2+6x+x+3-2-x^2+9=0}\\\sf{x^2+7x+10=0}$

$\sf{a=1~~b=7~~~c=10}\\\sf{\Delta=b^2-4ac}\\\sf{\Delta=7^2-4\cdot1\cdot10}\\\sf{\Delta=49-40}\\\sf{\Delta=9}\\\sf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\sf{x=\dfrac{-7\pm\sqrt{9}}{2\cdot1}}\\\sf{x=\dfrac{-7\pm3}{2}}\begin{cases}\sf{x_1=\dfrac{-7+3}{2}=-\dfrac{4}{2}=-2}\\\sf{x_2=\dfrac{-7-3}{2}=-\dfrac{10}{2}=-5}\end{cases}$