The number of real solutions of the equation $$\frac{(x+2)(x+3)(x+4)(x+5)}{(x-2)(x-3)(x-4)(x-5)} = 1$$ is

The denominator cannot be zero, so $$x\neq2,3,4,5$$.

We are given $$\dfrac{(x+2)(x+3)(x+4)(x+5)}{(x-2)(x-3)(x-4)(x-5)}=1$$.

Multiplying both sides by the denominator, we get $$ (x+2)(x+3)(x+4)(x+5)=(x-2)(x-3)(x-4)(x-5) $$.

We can pair the factors on both sides to get the same coefficient for $$x$$ as follows:

$$ (x+2)(x+5)=x^2+7x+10 $$

$$ (x+3)(x+4)=x^2+7x+12 $$.

Similarly,

$$ (x-2)(x-5)=x^2-7x+10 $$

$$ (x-3)(x-4)=x^2-7x+12 $$.

Therefore, the equation becomes $$ (x^2+7x+10)(x^2+7x+12)=(x^2-7x+10)(x^2-7x+12) $$.

Let $$ A=x^2+10 $$.

Then we get $$ (A+7x)(A+7x+2)=(A-7x)(A-7x+2) $$.

Expanding both sides gives $$ A^2+14Ax+49x^2+2A+14x=A^2-14Ax+49x^2+2A-14x $$.

Cancelling the common terms, we get $$ 28Ax+28x=0 $$.

Factoring, $$ 28x(A+1)=0 $$.

Since $$ A=x^2+10 $$, we get $$ 28x(x^2+11)=0 $$.

Therefore, $$ x(x^2+11)=0 $$.

So either $$ x=0 $$ or $$ x^2+11=0 $$.

But $$ x^2+11=0 $$ gives $$ x^2=-11 $$, which has no real solution because the square of a real number can never be negative.

Hence, the only real solution is $$ x=0 $$.

Therefore, the number of real solutions is $$ 1$$.