數學知識交流---解方程

一個青檸

2011-07-29 8:37 pm

(1) 利用一元二次方程公式，解方程 (x+1)x²+ (x-2)x - 3 = 0
(2) 解方程 px²+ qx + (p+q) = 0
(3) 利用公式 (x+y)(x-y) = x²- y²，解方程 ( x²+ 2x + 1 )( x²- 2x + 1 )( x⁴+ 2x²+ 1 ) + ( x²+ 1 )( x + 1 )( x - 1 ) + ( x²- 1 )( x²+ 1 ) + 1 = 0

回答 (1)

用戶2053

2011-07-30 2:13 am

✔ 最佳答案

1)(x+1)x² + (x-2)x - 3 = 0 x = [ -(x-2) ± √ ( (x-2)² - 4(x+1)(-3) ) ] / (2x+2)x = [ 2 - x ± √ (x² + 8x + 16) ] / (2x+2)x = [ 2 - x ± ±(x + 4) ] / (2x+2)x = [ 2 - x + (x + 4) ] / (2x+2) or x = [ 2 - x - (x + 4) ] / (2x+2)x² + x - 3 = 0 or x² + 2x + 1 = 0x = (- 1 ± √13)/2 or x = - 1
2) px² + qx + (p+q) = 0x = (- q ± √ (q² - 4p(p+q)) ) / (2p)
3)(x² + 2x + 1)(x² - 2x + 1)(x⁴+ 2x² + 1) + (x² +1)(x +1)(x -1) + (x² - 1)(x² + 1) + 1 = 0 ( (x² + 1)² - (2x)² )(x⁴+ 2x²+ 1) + (x² + 1)(x² - 1) + (x² - 1)(x² + 1) + 1 = 0(x⁴- 2x² + 1) (x⁴+ 2x² + 1) + 2 (x² + 1)(x² - 1) + 1 = 0(x² - 1)² (x² + 1)² + 2 (x² + 1)(x² - 1) + 1 = 0( (x² + 1)(x² - 1) + 1 )² = 0(x² + 1)(x² - 1) + 1 = 0x⁴= 0x = 0

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