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1)log(√(x²+1) + x) / log(√(x²+1) - x)= log(√(x²+1) + x) / log(√(x²+1) - x) + 1 - 1= [ log(√(x²+1) + x) + log(√(x²+1) - x) ] / log(√(x²+1) - x) - 1= log[(√(x²+1) + x) (√(x²+1) - x)] / log(√(x²+1) - x) - 1= (log 1) / log(√(x²+1) - x) - 1= - 1
2)y = 10 - √(x² - 10x + 29)
= 10 - √(x² - 10x + 25 + 4)
= 10 - √((x - 5)² + 4)when x = 5 , y(max) = 10 - √4 = 8
when x --> ∞ , y(min) --> - ∞
3)Let x items be he bought in the first visit :Old price for each - New price for each = $ 5/1210/x - 5/(x+5) = 5/122/x - 1/(x+5) = 1/122(x+5) - x = x(x+5) / 1224x + 120 - 12x = x² + 5xx² - 7x - 120 = 0(x - 15) (x + 8) = 0x = 15 or x = - 8(rejected)Answer : 15


2011-04-13 20:16:29 補充：
2)

when |x| --> ∞ , y(min) --> - ∞