Resolver $$ (x-9)(x-7)(x-5)(x-1)=(x-2)(x-4)(x-6)(x-10) $$
$$ (x-9)(x-7)(x-5)(x-1)=(x-2)(x-4)(x-6)(x-10)\\ [(x-9)(x-1)][(x-7)(x-5)]=[(x-2)(x-10)][(x-4)(x-6)]\\ (x^2-10x+9)(x^2-12x+35)=(x^2-12x+20)(x^2-10x+24)\\ $$ Haciendo $a=x^2-10x$ y $b=x^2-12x$ $$ (a+9)(b+35)=(b+20)(a+24)\\ ab+35a+9b+315=ba+24b+20a+480\\ 15a-15b=165\\ a-b=11 $$ Reemplazando $$ (x^2-10x)-(x^2-12x) = 11\\ 2x=11\\ \Rightarrow x=\frac{11}{2} $$