[Toán 9]Rút gọn

N

nguyentrachchung@gmail.com

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B

braga

Ta có: $z = \sqrt{2} + \sqrt[3]{3} \Leftrightarrow z - \sqrt{2} = \sqrt[3]{3}$

$\Leftrightarrow (z - \sqrt{2})^3 = 3 \Leftrightarrow z^3 + 6z - 3 = \sqrt{2}(3z^2 + 2)$



$\Leftrightarrow (z^3 + 6z - 3)^2 = 2(3z^2 + 2)^2$


$\Leftrightarrow z^6 - 6z^4 - 6z^3 - 6z^3 + 12z^2 - 36z + 1 = 0$


Khi đó:
$A = z(z^6 - 6z^4 - 6z^3 + 12z^2 - 36z + 1) + 2008 = 2008$
 
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