换元法 (x+1)(x+3)(x+5)(x+7)+15
问题描述:
换元法 (x+1)(x+3)(x+5)(x+7)+15
因式分解 (a-1)(a-2)(a-3)(a-4)-24
一道也可以!!!
答
(x+1)(x+3)(x+5)(x+7)+15
=(x+1)(x+7)(x+3)(x+5)+15
=(x²+8x+7)(x²+8x+15)+15
=(x²+8x)²+22(x²+8x)+15×7+15
=(x²+8x)²+22(x²+8x)+120
=(x²+8x+12)(x²+8x+10)
=(x+2)(x+6)((x²+8x+16)-6)
=(x+2)(x+6)(x+4+√6)(x+4-√6);
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