f(x)=1xx+12xx(x-1)x(x+1)3x(x-1)x(x-1)(x-2)(x-1)x(x+1)⇒f(2012)=

# $\mathrm{f}\left(\mathrm{x}\right)=$$\left|\begin{array}{ccc}1& x& x+1\\ 2x& x\left(x-1\right)& x\left(x+1\right)\\ 3x\left(x-1\right)& x\left(x-1\right)\left(x-2\right)& \left(x-1\right)x\left(x+1\right)\end{array}\right|$$⇒\mathrm{f}\left(2012\right)=$

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### Solution:

$\mathrm{f}\left(\mathrm{x}\right)=$$\left|\begin{array}{lcc}1& \mathrm{x}& \mathrm{x}+1\\ 2\mathrm{x}& \mathrm{x}\left(\mathrm{x}-1\right)& \mathrm{x}\left(\mathrm{x}+1\right)\\ 3\mathrm{x}\left(\mathrm{x}-1\right)& \mathrm{x}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)& \left(\mathrm{x}-1\right)\mathrm{x}\left(\mathrm{x}+1\right)\end{array}\right|$

$f\left(x\right)={x}^{2}\left(x-1\right)\left(x+1\right)$

$f\left(x\right)={x}^{2}\left(x-1\right)\left(x+1\right)$  +91

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