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[boykovipk602]

Cho $a=\frac{1-\sqrt{2}}{2}$. Tính $\sqrt{16a^{8}-51a}$

[Nguyen Minh Hai]

Ta có: $\sqrt{16a^8-51a}=\sqrt{\frac{(2a)^8-8.51.2a}{16}}$

$2a=1-\sqrt{2}$

$(2a)^2=3-2\sqrt{2}$  

$\Leftrightarrow (2a)^4=(3-2\sqrt{2})^2=17-12\sqrt{2}$  

$\Leftrightarrow (2a)^8=(17-12\sqrt{2})^2=577-408\sqrt{2}$

$\sqrt{16a^8-51a}=\sqrt{\frac{577-408\sqrt{2}-8.51.(1-\sqrt{2})}{16}}=\sqrt{\frac{169}{16}}=\frac{13}{4}$

Edited by Nguyen Minh Hai, 23-11-2014 - 23:32.