Answer:
The sum of the given expressions is [tex]\frac{(2+x)(8-x)}{x(x+4)}[/tex]
x = 0 and x = -4
Step-by-step explanation:
we need to find the sum of the given expressions
[tex]\frac{6}{4+x} +\frac{4-x}{x} \\= \frac{6*x+(4-x)(4+x)}{(4+x)*x} \\=\frac{6x+16-x^2}{x(x+4)} \\=\frac{16+6x-x^2}{x(x+4)} \\=\frac{(2+x)(8-x)}{x(x+4)}[/tex]
To find excluded values:
x ( 4 + x ) = 0
x = 0 and 4 + x = 0
∴ x = 0 and x = -4
