急急問:再請問附圖的三題如何解呢？謝謝喔～?

1)
(a - 2√(ab) + b)^½ + (a + 2√(ab) + b)^½ , 因 b > a , 故 (a - 2√(ab) + b)^½ = √b - √a ,
= (√b - √a) + (√b + √a)
= 2√b

2)
f(2) = a + b 2ⁿ = 17
f(4) = a + b 4ⁿ = 77
f(8) = a + b 8ⁿ = 377
則 b(8ⁿ - 4ⁿ) / (b(4ⁿ - 2ⁿ)) = (377 - 77) / (77 - 17)
4ⁿ(2ⁿ - 1) / 2ⁿ(2ⁿ - 1) = 5
2ⁿ = 5
故 a + 5b = 17 ; a + 25b = 77 ; a + 125b = 377 , 解得 a = 2 , b = 3 ⇒ a² + b² = 13.

3)
[10²ºº¹ / (10⁶⁶⁷ + 2002)]
= [ ((10⁶⁶⁷)³ + 2002³ ) / (10⁶⁶⁷ + 2002) - 2002³ / (10⁶⁶⁷ + 2002) ]
因 2002³ / (10⁶⁶⁷ + 2002) < 1 , 原式
= ((10⁶⁶⁷)³ + 2002³ ) / (10⁶⁶⁷ + 2002) - 1
= (10⁶⁶⁷)² - (10⁶⁶⁷)2002 + 2002² - 1
= (10¹³³º - (10⁶⁶³)2002) × 10000 + 2002² - 1
原式末四位數 = 2002² - 1 = 2001 × 2003 = 2000² + (1+3)2000 + 3 = 2000² + 8003 的末四位數 = 8003.