Answer:
Solution given:
[tex]f(x)=\frac{x-16}{x^2+6x-40}[/tex]
[tex]g(x)=\frac{1}{x+10}[/tex]
now
f(x)+g(x)=[tex]\frac{x-16}{x^2+6x-40}+\frac{1}{x+10}[/tex]....(1)
now
factoring x²+6x-40
we get
x²+10x-4x-40
x(x+10)-4(x+10)
(x+10)(x-4)
now substituting in equation 1 ,we get
f(x)+g(x)=[tex]\frac{x-16}{(x+10)(x-4)}+\frac{1}{x+10}[/tex]
taking l.c.m
=[tex]\frac{(x-16)+(x-4)}{(x-10)(x-4)}[/tex]
=now
opening bracket
[tex]\frac{x-16+x-4}{x²-10x-4x+40}[/tex]
=[tex]\frac{2x-20}{x²+6x-40}[/tex]
So
answer is :
B. [tex]\bold b\frac{2x-20}{x^2+6x-40}[/tex]
