Question

$\bold{\boxed{Answer it fast(12 th part)}}$​

Answer 1

           

$Let \ \ n-10=k \\ \\ \therefore\ 10-n=-n+10=-(n-10)=-k \\ \\ \\ (n-10)^2+(10-n) \\ \\ \Rightarrow\ k^2-k \\ \\ \Rightarrow\ k(k-1) \\ \\ \Rightarrow\ (n-10)(n-10-1) \\ \\ \Rightarrow\ (n-10)(n-11) \\ \\ \\ \therefore\ (n-10)^2+(10-n)=(n-10)(n-11)$

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