Question

!!!!!!!!!!!!!!!!por favor ​

Answer 1

$\frac{2x}{x-1}+\frac{3}{x-3}=\frac{x+3}{(x-1)\cdot (x-3)}\\\\\frac{2x\cdot (x-3)}{(x-1)\cdot (x-3)}+\frac{3 \cdot (x-1)}{(x-1)\cdot (x-3)}=\frac{x+3}{(x-1)\cdot (x-3)}\\\\\frac{2x^2-6x}{(x-1)\cdot (x-3)}+\frac{3x-3}{(x-1)\cdot (x-3)}=\frac{x+3}{(x-1)\cdot (x-3)}\\\\\frac{2x^2-6x+3x-3}{(x-1)\cdot (x-3)}=\frac{x+3}{(x-1)\cdot (x-3)}\\\\2x^2-6x+3x-3=x+3\\\\2x^2-6x+3x-3-x-3=0\\\\2x^2-4x-6=0 \ \ \Longrightarrow (:2)\\\\x^2-2x-3=0$

$\\x^2-2x-3=0\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-2)\pm \sqrt{(-2)^2-4 \cdot 1 \cdot (-3)}}{2\cdot 1}=\frac{2\pm \sqrt{4+12}}{2}=\frac{2\pm \sqrt{16}}{2}= \frac{2\pm 4}{2}\\\\x^'= \frac{2+4}{2}=\frac{6}{2}=3 \ \ \ \ \ \ x^''= \frac{2-4}{2}=\frac{-2}{2}=-1$

Observe que:

        x - 1 ≠ 0   e  x - 3 ≠ 0

logo, x ≠ 1        e   x ≠ 3

Devido a esse fato, excluímos x = 3, e teremos como solução x = -1

S = {-1}

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

$\frac{x+7}{x+2}+\frac{x-3}{x-2}=5\\\\\frac{(x+7) \cdot (x-2)}{(x+2)\cdot (x-2)}+\frac{(x-3)\cdot(x+2)}{(x+2) \cdot(x-2)}=5\\\\\frac{x^2-2x+7x-14}{x^2-4}+\frac{x^2+2x-3x-6}{x^2-4}=5\\\\\\\frac{x^2+5x-14}{x^2-4}+\frac{x^2-x-6}{x^2-4}=5\\\\\frac{x^2+5x-14+x^2+x^2-x-6}{x^2-4}=5\\\\\\\frac{x^2+x^2+5x-x-14-6}{x^2-4}=5\\\\\frac{2x^2+4x-20}{x^2-4}=5\\2x^2+4x-20=5 \cdot (x^2-4)\\2x^2+4x-20=5x^2-20\\2x^2-5x^2+4x-20+20=0\\-3x^2+4x=0 \ \Longrightarrow (:x) \Rightarrow x=0 \\-3x+4=0$

$-3x=-4 x=\frac{-4}{-3}\\\\x=\frac{4}{3}$

S = {0, 4/3}