We note that $361=19^2$ and $441=21^2$, so $x=21^2+2(21)(19)+19^2$. This is just the binomial expansion of $(21+19)^2=40^2=\boxed{1600}$.
Question:
$441+2(21)(19)+361=x$. Solve for $x$.
Answer:
Let's think step by step.

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1600
$441+2(21)(19)+361=x$. Solve for $x$.
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