[Toán 8] Các số thực

T

tukithansau_1999

H

hiensau99

$x^{2} + y^{2} = 1\to x^4+y^4=1-2x^2y^2$
$\dfrac{x^4}{a} + \dfrac{y^4}{b} = \dfrac{x^2+y^2}{a+b} $

$\to \dfrac{x^4b+y^4a}{ab}= \dfrac{x^2+y^2}{a+b}$

$\to (x^4b+y^4a)(a+b)=ab(x^2+y^2)$

$\to (x^4+y^4)ab+x^4b^2+y^4a^2-ab=0 $

$\to (1-2x^2y^2)ab+x^4b^2+y^4a^2-ab=0$

$\to x^4b^2+y^4a^2-2x^2ay^2b=0$

$\to (x^2b-y^2a)^2=0$

$\to x^2b=y^2a \to \dfrac{x^2}{a}= \dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b} $

$\to \dfrac{x^{2006}}{a^{1003}} = \dfrac{y^{2006}}{b^{1003}} = \dfrac{1}{(a+b)^{1003}}$

$\to \dfrac{x^{2006}}{a^{1003}} + \dfrac{y^{2006}}{b^{1003}} = \dfrac{2}{(a+b)^{1003}}$
 
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