b) Ta có
$\dfrac{(x-2)!}{(x-4)!} + \dfrac{x!}{2! (x-2)!}= 101$
$<-> (x-2)(x-3) + \dfrac{x(x-1)}{2} = 101$
$<-> 2(x^2 - 5x + 6) + x^2 - x = 202$
$<-> 3x^2 -11x -190 = 0$
$<-> x = 10$
c) Ptrinh tương đương vs
$\dfrac{x!}{(x-1)!} + 6\dfrac{x!}{2! (x-2)!} + 6 \dfrac{x!}{3!(x-3)!} = 9x^2 - 28$
$<-> x + 3x(x-1) + x(x-1)(x-2) = 9x^2 - 28$
$<-> x^3 -9x^2 + 28 = 0$
$<-> (x-2)(x^2 -7x -14) = 0$
Vậy $x = 2$, do đó vô nghiệm.
