Question

me ayudan porfa es importante

Answer 1

Respuesta:

Un gusto ayudarte

$\frac{x(x+5)}{4x + 4} =-\frac{9}{5} \\\\5({x(x+5)})=-(4x + 4){9}\\\\5(x^{2} +5x) =-36x +36\\\\5x^{2} +25x = -36x -36\\\\5x^{2} +25x + 36x= -36\\\\5x^{2} +61x +36 = 0$

$\mathrm{Para\:}\quad a=5,\:b=61,\:c=36$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x_{1,\:2}=\frac{-61\pm \sqrt{61^2-4\cdot \:5\cdot \:36}}{2\cdot \:5}$

$x_{1,\:2}=\frac{-61\pm \sqrt{3001}}{2\cdot \:5}$

$x_1=\frac{-61+\sqrt{3001}}{10}$

$\:x_2=\frac{-61-\sqrt{3001}}{10}$

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$\frac{x-1}{x+2}=\frac{10}{3x-2} \\\\({3x-2})({x-1}) = ({x+2})({10}) \\\\3x^2-5x+2 = 10x+20 \\\\3x^2-5x- 10x=+20-2\\\\3x^2-15x=18\\\\3x^2-15x-18=0\\\\\frac{3x^2}{3} -\frac{15x}{3} -\frac{18}{3} =\frac{0}{3} \\\\x^{2} -5x - 6 =0\\\\(x-6)(x-1)=0\\\\x_{1} -6 =0\\\\x_{1} =6\\\\\\\\x_{2} -1 =0\\\\x_{2} =1\\\\x_{1,2} =(6,1)$

Saludos desde Ecuador "ya llegara el dia bro :) no apresures las cosas "