求解二次方程(中一下)問題

1)1/(x-1) + 1/(x-4) = 1/(x-2) + 1/(x-3)(x-4 + x-1) / ( (x-1)(x-4) ) = (x-3 + x-2) / ( (x-2)(x-3) ) (2x-5) / ( (x-1)(x-4) ) = (2x-5) / ( (x-2)(x-3) ) (2x-5)(x-2)(x-3) = (2x-5)(x-1)(x-4) (2x - 5) (x² - 5x + 6) = (2x - 5) (x² - 5x + 4)(2x - 5) (x² - 5x + 6 - x² + 5x - 4) = 0(2x - 5) (2) = 02x - 5 = 0x = 5/2
2)x + 1/x = - 4(x + 1/x)² = 16x² + 2x(1/x) + 1/x² = 1 6x² - 2x(1/x) + 1/x² = 16 - 4 x(1/x)x² - 2x(1/x) + 1/x² = 12(x - 1/x)² = 12x - 1/x = ± √12 x - 1/x = 2√3 或 - 2√3
3)設此商人原先買進芒果 x 公斤 , 則每公斤成本價 6000/x 元。淘汰賣相不佳的芒果餘下(x - 30) 公斤 ,
以 6000/x + 10 元賣出得款 8100 元 :(x - 30) (6000/x + 10) = 8100(x - 30) (6000 + 10x) = 8100x(x - 30) (x + 600) = 810x x² + 570x - 18000 = 810x x² - 240x - 18000 = 0(x - 300) (x + 60) = 0x = 300 或 x = - 60 (捨)此商人原先買進芒果 300 公斤。

2012-03-02 21:39:04 補充：
以每公斤 6000/x + 10 元賣出得款 8100 元