มัธยมต้น ตอนที่ 1
อ้างอิง:
12. กำหนดให้ $\dfrac{x^2-y^2}{x^2+y^2}+\dfrac{x^2+y^2}{x^2-y^2} = k$ จงหาค่าของ $\dfrac{x^8-y^8}{x^8+y^8}+\dfrac{x^8+y^8}{x^8-y^8}$ ในรูปของ $k$
(เสนอโดย คุณ -InnoXenT-)
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$\dfrac{x^2-y^2}{x^2+y^2}+\dfrac{x^2+y^2}{x^2-y^2} = k$
$(x^2+y^2)^2 + (x^2-y^2)^2 = k(x^2-y^2)(x^2+y^2)$
$(x^4+y^4+2x^2y^2) +(x^4+y^4-2x^2y^2) = k(x^4-y^4)$
$2x^4+2y^4 = kx^4-ky^4$
$ 2x^4-kx^4 = -ky^4-2y^4$
$x^4(2-k) = -(2+k)y^4$
$x^4 = \frac{k+2}{k-2}y^4$
แทนค่า x ใน$\dfrac{x^8-y^8}{x^8+y^8}+\dfrac{x^8+y^8}{x^8-y^8}$
$ = \dfrac{( \frac{k+2}{k-2}y^4)^2 - y^8}{(\frac{k+2}{k-2}y^4)^2 + y^8} + \dfrac{(\frac{(k+2}{k-2}y^4)^2 + y^8}{(\frac{k+2}{k-2}y^4)^2 - y^8} $
$ = \dfrac{( \frac{k+2}{k-2})^2 - 1}{(\frac{k+2}{k-2})^2 + 1} + \dfrac{(\frac{k+2}{k-2})^2 + 1}{(\frac{k+2}{k-2})^2 - 1} $
$\dfrac{\frac{k^2+4k+4}{k^2-4k+4}-1}{\frac{k^2+4k+4}{k^2-4k+4}+1} +
\dfrac{\frac{k^2+4k+4}{k^2-4k+4}+1}{\frac{k^2+4k+4}{k^2-4k+4}-1}$
$ = \dfrac{k^2+4k+4-k^2+4k-4}{k^2+4k+4+k^2-4k+4} + \dfrac{k^2+4k+4+k^2-4k+4}{k^2+4k+4-k^2+4k-4}$
$ = \dfrac{4k}{k^2+4} + \dfrac{k^2+4}{4k}$
$= \dfrac{k^4+24k^2+16}{4k(k^2+4)}$