Question

AYÚDEMEN CONE STO EJERCICIO POR FAVOR​

Answer 1

Respuesta:

Explicación paso a paso:

a)

$f(x) + g(x) = (x^2-1)+(2x+3)=x^2+2x+2$

b)

$f(x) + h(x) =(x^2-1)+(\frac{2}{x+1} )=\frac{(x^2-1)*(x+1)+2}{x+1} =\frac{x^3+x^2-x-1+2}{x+1} =\frac{x^3+x^2-x+1}{x+1}$

c)

$h(x) - g(x) =(\frac{2}{x+1} )-(2x+3)=\frac{2-(2x+3)(x+1)}{x+1}=\frac{2-(2x^2+2x+3x+3)}{x+1}=\frac{2-2x^2-5x-3}{x+1} =-\frac{2x^2+5x+1}{x+1}$

d)

$g(x)-h(x)=(2x+3)-(\frac{2}{x+1} )=\frac{(2x+3)(x+1)-2}{x+1} =\frac{(2x^2+2x+3x+1)-2}{x+1}=\frac{2x^2+5x-1}{x+1}$

e)

$f(x).g(x)=(x^2-1).(2x+3)=2x^3+3x^2-2x-3$

f)

$g(x).h(x)=(2x+3).(\frac{2}{x+1})=\frac{2.(2x+3)}{x+1} =\frac{4x+6}{x+1}$

g)

$\frac{f(x)}{g(x)}=\frac{x^2-1}{2x+3}$

h)

$\frac{f(x)}{h(x)} =\frac{x^2-1}{\frac{2}{x+1} } =\frac{(x^2+1)(x+1)}{2} =\frac{x^3+x^2+x+1}{2}$