An algebra problem by Ash R

$\begin{aligned} x^2 & = \frac {19!+20!+21!}{19!} & \small \color{#3D99F6} \text{Let }a = 20 \\ & = \frac {(a-1)!+a!+(a+1)!}{(a-1)!} \\ & = \frac {\cancel{(a-1)!}+a\cancel{(a-1)!}+(a+1)a\cancel{(a-1)!}}{\cancel{(a-1)!}} \\ & = 1 + a + a(a+1) \\ & = a^2 + 2a + 1 \\ & = (a+1)^2 \\ \implies x & = a+1 = \boxed{21} \end{aligned}$

We have

$x^2= \dfrac{19!+20!+21!}{19!} = \frac{19!}{19!} + \frac{20!}{19!} +\frac{21!}{20!} =1+20+20×21 = 21^2$

Therefore $x= 21$

$\dfrac{19!+20!+21!}{19!}=x^2$

Bring out the common factor of the numerator. We have

$\dfrac{19!\left[1+20+20(21)\right]}{19!}=x^2$

$19!$ cancels out, so

$21+420=x^2$

$441=x^2$

$\sqrt{441}=x$

$\boxed{21=x}$

(19! + 20! + 21!) / 19!

=(19! + 19!×20 + 19!×20×21) / 19

= 1 + 20 + 20×21 = 441

X^2=441

X=21