(x-1)(x-3)(x-5)(x-7)-9 因式分解,要用换元法

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(x-1)(x-3)(x-5)(x-7)-9 因式分解,要用换元法
(x-1)(x-3)(x-5)(x-7)-9 因式分解,要用换元法

(x-1)(x-3)(x-5)(x-7)-9 因式分解,要用换元法
解 原式=(x-1)(x-7)(x-3)(x-5)-9
=(x^2-8x+7)(x^2-8x+15)
设x^2-8x+7=a 则
原式=a(a+8)-9
=a^2+8a-9
=(a-1)(a+9)
=(x^2-8x+6)(x^2-8x+16)

(x-1)((x-1)-2)((x-1)-4)((x-1)-6)-9
=(x-1)(1-2-4-6)

(x-1)(x-3)(x-5)(x-7)-9
=(x-1)(x-7)(x-3)(x-5)-9
=(x²-8x+7)(x²-8x+15)-9
=(x²-8x)²+22(x²-8x)+105-9
=(x²-8x)²+22(x²-8x)+96
=(x²-8x+6)(x²-8x+16)
==(x²-8x+6)(x-4)²


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