Respuesta :
we have the problem
[tex]\frac{4x}{x^2+x-6}=\frac{7x}{x^2-5x-24}+\frac{3}{x^2-10x+16}[/tex]Simplify the denominators
so
x^2+x-6=(x+3)(x-2)
x^2-5x-24=(x+3)(x-8)
x^2-10x+16=(x-2)(x-8)
sustitute en la expresion original
[tex]\frac{4x}{\mleft(x+3\mright)\mleft(x-2\mright)}=\frac{7x}{\mleft(x+3\mright)\mleft(x-8\mright)}+\frac{3}{\mleft(x-2\mright)\mleft(x-8\mright)}[/tex]Multiplica ambos lados por (x+3)(x-8)(x-2) para eliminar fracciones
[tex]\frac{4x\mleft(x+3\mright)\mleft(x-8\mright)\mleft(x-2\mright)}{(x+3)(x-2)}=\frac{7x\mleft(x+3\mright)\mleft(x-8\mright)\mleft(x-2\mright)}{(x+3)(x-8)}+\frac{3\mleft(x+3\mright)\mleft(x-8\mright)\mleft(x-2\mright)}{(x-2)(x-8)}[/tex]simplifica
[tex]4x(x-8)=7x(x-2)+3(x+3)[/tex][tex]4x^2-32x=7x^2-14x+3x+9[/tex][tex]\begin{gathered} 4x^2-32x=7x^2-11x+9 \\ 7x^2-4x^2+32x-11x+9=0 \\ 3x^2+21x+9=0 \end{gathered}[/tex]simplifica, divide por 3 toda la expression
[tex]x^2+7x+3=0[/tex]Resuelve la ecuacion quadratica utilizando la formula
a=1
b=7
c=3
sustituye
[tex]x=\frac{-7\pm\sqrt[]{7^2-4(1)(3)}}{2(1)}[/tex][tex]x=\frac{-7\pm\sqrt[]{27}}{2}[/tex]Las soluciones son
[tex]x=\frac{-7+\sqrt[]{27}}{2}[/tex][tex]x=\frac{-7-\sqrt[]{27}}{2}[/tex]