Question

X/(x square -25) - 8/(x square +6x+5) can be written as

Answer 1

Step-by-step explanation:

x/x²-25 x/(x+5)(x-5)

[8/x²+6x+5]=[8/x²+5x+x+5]=[8/(x+5)(x+1)]

find lcm of denominators which are (x+5)(x-5) & (x+5)(x+1)

lcm=(x+5)(x-5)(x+1)=(x²-25)(x+1)=x³+x²-25x-25

x(x+1)-8(x-5)/x³+x²-25x-25

x²+x-8x+40/x³+x²-25x-25

x²-7x+40/x³+x²-25x-25

Answer 2

The final answer is: $\frac{x^2-7x+40}{(x^3+x^2-25x-25)}$

Step-by-step explanation:

$\ Given \ value:\\\\\frac{x}{x^2-25} - \frac{8}{x^2+6x+5} \\\\\ Solution:\\\\\frac{x}{x^2-25} - \frac{8}{x^2+6x+5} \\\\\therefore x^2+6x+5 \\\\\rightarrow x^2+(5+1)x+5 \\\\ \rightarrow x^2+5x+1x+5 \\\\ \rightarrow x(x+5)+1(x+5) \\\\ \rightarrow (x+5)(x+1) \\\\ \rightarrow \frac{x}{(x^2-5^2)} - \frac{8}{(x+5)(x+1)} \\\\\rightarrow \frac{x}{(x-5)(x+5)} - \frac{8}{(x+5)(x+1)} \\\\\rightarrow \frac{x(x+1) - 8(x-5)}{(x-5)(x+5)(x+1)}\\\\\rightarrow \frac{x^2+x - 8x+40}{(x-5)(x+5)(x+1)}$

$\rightarrow \frac{x^2-7x+40}{(x-5)(x+5)(x+1)}\\\\\rightarrow \frac{x^2-7x+40}{(x^2-25)(x+1)}\\\\\rightarrow \frac{x^2-7x+40}{(x^3+x^2-25x-25)}$

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