How do you simplify $((x^2-25 )/(x^2+6+5))/(x / x^2)$ ?

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Answer 1 · 2016-07-27T12:06:42

$(x(x-5))/((x+1))$

In this form, the fraction just looks nasty!!

We can write $(color(red)a/color(blue)b)/(color(blue)c/color(red)d)$ in the much easier form of $color(red)(axxd)/color(blue)(bxxc)$

Let's do the same for $(color(red)(x^2-25 )/color(blue)(x^2+6+5))/color(blue)(x / color(red)(x^2))$

= $color(red)((x^2-25 )xx x^2)/color(blue)((x^2+6+5)xx x)" Much better!"$

Now factorise and cancel like factors.

= $color(red)(((x+5)(x-5)xx x^2)/color(blue)((x+5)(x+1)xx x)$

= $(cancel(x+5)(x-5)xx x^cancel2)/(cancel(x+5)(x+1)xx cancelx)$

= $(x(x-5))/((x+1))$

How do you simplify ((x^2-25 )/(x^2+6+5))/(x / x^2)((x^2-25 )/(x^2+6+5))/(x / x^2)?