Prove that f(5)*f(6)*f(7)*f(8)*f(9)*f(10)=7 if f(x)=(x^2+x-12)/(x^2-16).

Before calculating the product
f(5)*f(6)*f(7)*f(8)*f(9)*f(10)=7, we'll try to simplify the ratio
(x^2+x-12)/(x^2-16).

We notice that the denominator is a
difference of squares so we'll re-write it as a
product:

a^2 - b^2 =
(a-b)(a+b)

We'll put a^2 = x^2 and b^2 =
16

x^2 - 16 = (x-4)(x+4)

Now,
we'll try to factorize the numerator. For this reason, we'll compute the roots of the
expression x^2+x-12.

x^2+x-12 =
0

We'll apply the quadratic
formula:

x1 = [-1 + sqrt(1 +
48)]/2

x1 = (-1+7)/2

x1 =
3

x2 = (-1-7)/2

x2 =
-4

Now, we'll factorize the
expression:

x^2+x-12 =
(x-x1)(x-x2)

We'll substitute x1 and
x2:

x^2+x-12 = (x-3)(x+4)

Now,
we'll re-write f(x):

f(x) =
(x-3)(x+4)/(x-4)(x+4)

We'll simplify and we'll
have:

f(x) = (x-3)/(x-4)

Now,
we'll plug in values for x:

x = 5 => f(5) =
(5-3)/(5-4)

              f(5) =
2/1

x = 6 => f(6) =
(6-3)/(6-4)

              f(6) =
3/2

x = 7 => f(7) =
(7-3)/(7-4)

              f(7) =
4/3

x = 8 => f(8) =
(8-3)/(8-4)

              f(8) =
5/4

x = 9 => f(9) =
(9-3)/(9-4)

              f(9) =
6/5

x = 10 => f(10) =
(10-3)/(10-4)

              f(10) =
7/6

Now, we'll calculate the
product:

f(5)*f(6)*f(7)*f(8)*f(9)*f(10)=(2/1)(3/2)(4/3)(5/4)(6/5)(7/6)

We'll simplify
like terms and we'll get:

f(5)*f(6)*f(7)*f(8)*f(9)*f(10) =
7/1

f(5)*f(6)*f(7)*f(8)*f(9)*f(10) =
7