初中因式分解换元法例题 初二上册代数题，请用换元法来做 因式分解 题目如下
\u521d\u4e8c\u7684\u56e0\u5f0f\u5206\u89e3\uff0c\u7528\u6362\u5143\u6cd5\u8ba1\u7b97\u3002\u7b54\u5bf9\u7ed9\u5206\u3002\u8bbe y = (x² + 2x - 1)
(x² + 2x - 1)² + 2(x² + 2x) - 10
= (x² + 2x - 1)² + 2(x² + 2x - 1) - 8
= y² + 2y - 8
= (y - 2)(y + 4)
= (x² + 2x - 3)(x² + 2x + 1)
= (x + 3)(x - 1)(x + 1)²
\u4e0b\u4e00\u9898\u53ea\u8981\u6bcf\u9879\u4e582\u5c31\u597d\u4e86:
½ x² + x + ½ = 0
x² + 2x + 1 = 0
(x + 1)² = 0
x = -1
\u89e3\uff1a
(x+1)(x+2)(x+3)(x+4)-24
=[(x+1)(x+4)][(x+2)(x+3)]-24
=\ufe59x²\uff0b5x\uff0b4)(x²\uff0b5x\uff0b6\ufe5a-24
\u8bbex²+5x=a\uff0c\u6362\u5143\u5f97
\u539f\u5f0f
=(a+4)(a+6)-24
=a²+4a+6a+24-24
=a²+10a
=a(a+10)
=(x²\uff0b5x)(x²\uff0b5x\uff0b10)
=x(x+5)(x²\uff0b5x\uff0b10)
=[(x+1)(x+4)][(x+2)(x+3)]-24
=﹙x²＋5x＋4)(x²＋5x＋6﹚-24
设x²+5x=a,换元得
原式
=(a+4)(a+6)-24
=a²+4a+6a+24-24
=a²+10a
=a(a+10)
=(x²＋5x)(x²＋5x＋10)
=x(x+5)(x²＋5x＋10)
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