Question

مطلوب است محاسبه$$ P( \sqrt{2} )$$ :

$$P(x+ \frac{2}{x} )= x^{12} + x^{6} +1$$ $$\Longrightarrow x+ \frac{2}{x} = \sqrt{2}, x^{2} = \sqrt{2} x-2,... $$

Answer 1

$x+ \frac{2}{x}= \sqrt{2} \Rightarrow x.x+ x.\frac{2}{x} =x. \sqrt{2} \Rightarrow x^2+2= \sqrt{2} x\Rightarrow x^2= \sqrt{2} x-2$

$ \Rightarrow x.x^2=x(\sqrt{2} x-2) \Rightarrow x^3= \sqrt{2} x^2-2x= \sqrt{2} ( \sqrt{2} x-2)-2x=2x-2 \sqrt{2} -2x$

$ \Rightarrow x^3=-2 \sqrt{2} \Rightarrow x^6=8,x^{12}=64$

$P( \sqrt{2} )=64+8+1=73$

$ \Box $