对下列题目进行因式分解:1.(x+1)*(x+2)*(x+3)*(x+4)-120 2.(x-1)*(x-3)*(x+1
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对下列题目进行因式分解:1.(x+1)*(x+2)*(x+3)*(x+4)-120 2.(x-1)*(x-3)*(x+1)*(x+3)-20
括号之间的是乘号
用拆项添项法
求求求
括号之间的是乘号
用拆项添项法
求求求
![对下列题目进行因式分解:1.(x+1)*(x+2)*(x+3)*(x+4)-120 2.(x-1)*(x-3)*(x+1](/uploads/image/z/19817865-9-5.jpg?t=%E5%AF%B9%E4%B8%8B%E5%88%97%E9%A2%98%E7%9B%AE%E8%BF%9B%E8%A1%8C%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%EF%BC%9A1.%EF%BC%88x%2B1%EF%BC%89%2A%EF%BC%88x%2B2%EF%BC%89%2A%EF%BC%88x%2B3%EF%BC%89%2A%EF%BC%88x%2B4%EF%BC%89-120+2.%EF%BC%88x-1%EF%BC%89%2A%EF%BC%88x-3%EF%BC%89%2A%EF%BC%88x%2B1)
1.(x+1)*(x+2)*(x+3)*(x+4)-120 =(x+1)(x+4)(x+2)(x+3)-120=(x^2+5x+4)(x^2+5x+6)-120
设t=x^2+5x则原式=(t+4)(t+6)-120=t^2+10t+24-120=t^2+10t-96(十字相乘)=(t+16)(t-6)=(x^2+5x+16)(x^2+5x-6)=(x^2+5x+16)(x+6)(x-1)
2..(x-1)*(x-3)*(x+1)*(x+3)-20=(x^2+3-4x)(x^2+3+4x)-20=(x^2+3)^2-16x^2-20=x^4+6x^2+9-20-16x^2=x^4-10x^2-11=(x^2-11)(x^2+1)
=(x-根号11)(x+根号11)(x^2+1)
设t=x^2+5x则原式=(t+4)(t+6)-120=t^2+10t+24-120=t^2+10t-96(十字相乘)=(t+16)(t-6)=(x^2+5x+16)(x^2+5x-6)=(x^2+5x+16)(x+6)(x-1)
2..(x-1)*(x-3)*(x+1)*(x+3)-20=(x^2+3-4x)(x^2+3+4x)-20=(x^2+3)^2-16x^2-20=x^4+6x^2+9-20-16x^2=x^4-10x^2-11=(x^2-11)(x^2+1)
=(x-根号11)(x+根号11)(x^2+1)
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