請問: 設(√11+√√7)^2=n+a，其中n為正整數， 0<a<1，為了求出n之值，可令x=√11+√7， y=√11-√7，則: (1)x^2+y^2=? (2)x^4+y^4=? (3)n=? (4)請問為何要令x=√11+√7，y=√11-√7呢？ 謝謝~?

(√11+√√7)^2=n+a，n=正整數， 0<a<1，令x=√11+√7， y=√11-√7，則:


(1) x^2 + y^2 = ?

= (√11+√7)^2 + (√11-√7)^2

= (11 + 2√77 + 7) + (11 - 2√77 + 7)

= 22 + 14

= 36




(2) x^4 + y^4 = ?


(2a) x + y = ?

= (√11 + √7) + (√11 - √7)

= 2√11


(2b) xy = ?

(x + y)^2

= (x^2 + y^2) + 2xy

= 36 + 2xy ;;; by (1)

= 4*11 ;;; by (2a)

= 44


=> xy = (44-36)/2 = 4


(2c) x^4 + y^4

= (x^2 + y^2)^2 - 2(xy)^2

= 36^2 - 2*16

= 1296 - 32

= 1264



(3) n = ?

n + a = (√11 + √√7)^2

= 11 + √7 + 2√11*√√7

= 11 + [√7 + 2√√(121*7)]

= 11 + (√7 + 2√√847)

= 11 + (2.65 + 2*5.395)

= 11 + 2 + 10 + (0.65 + 0.79)

= 23 + 1.44

= 24 + 0.44



=> n = 24

=> a = 0.44



(4) 為何要令: x = √11 + √7， y = √11 - √7 ?

Ans: 憑多做題目所產生的經驗與妙智慧出來