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shepuryov [24]
3 years ago
13

Add the given expressions, simplifying each result and noting the combined excluded values.

Mathematics
1 answer:
algol [13]3 years ago
8 0

Answer:

The sum of the given expressions is \frac{(2+x)(8-x)}{x(x+4)}

x = 0 and x = -4

Step-by-step explanation:

we need to find the sum of the given expressions

\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)}

To find excluded values:

x ( 4 + x ) = 0

x = 0  and 4 + x = 0

∴ x = 0 and x = -4

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