Question

ayuda porfa es urgente ​

Answer 1

Respuesta:

[tex]=\frac{x^2+20}{\left(x-2\right)\left(x+3\right)\left(x+5\right)}[/tex]

Explicación paso a paso:

   [tex]\frac{x+5}{x^2+x-6}-\frac{x+3}{x^2+3x-10}+\frac{x-2}{x^2+8x+15}[/tex]

[tex]=\frac{x+5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{x^2+3x-10}+\frac{x-2}{x^2+8x+15}[/tex]

[tex]=\frac{x+5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+5\right)}+\frac{x-2}{\left(x+3\right)\left(x+5\right)}[/tex]

Entonces, reescribimos las fracciones basándonos en el mínimo común denominador:

[tex]=\frac{\left(x+5\right)^2}{\left(x-2\right)\left(x+3\right)\left(x+5\right)}-\frac{\left(x+3\right)^2}{\left(x-2\right)\left(x+3\right)\left(x+5\right)}+\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)\left(x+5\right)}[/tex]

[tex]=\frac{\left(x+5\right)^2-\left(x+3\right)^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)\left(x+5\right)}[/tex]

Expandimos:

[tex]=\frac{x^2+20}{\left(x-2\right)\left(x+3\right)\left(x+5\right)}[/tex]