因式分解:1.2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²2.(x+1)(x+3)(x+5)(x+7)+15看答案看了半天也没搞懂,唉,我初高中脱节了

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 17:43:50

因式分解:1.2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²2.(x+1)(x+3)(x+5)(x+7)+15看答案看了半天也没搞懂,唉,我初高中脱节了
因式分解:
1.2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²
2.(x+1)(x+3)(x+5)(x+7)+15
看答案看了半天也没搞懂,唉,我初高中脱节了

因式分解:1.2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²2.(x+1)(x+3)(x+5)(x+7)+15看答案看了半天也没搞懂,唉,我初高中脱节了
第一题(2(x²+6x+1)+(x²+1))*((x²+6x+1)+2(x²+1))
第二题 ((x+1)(x+7))((x+3)(x+5))+15
=(x²+8x+7)(x²+8x+15)+15
=(x²+8x+7)(x²+8x+15)+8(x²+8x+15)-8(x²+8x+15)+15
=(x²+8x+15)²-8(x²+8x+15)+15
=(x²+8x+15-3)(x²+8x+15-5)

1
2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²
=(2(x²+6x+1)+(x²+1))((x²+6x+1)+2(x²+1))
2
(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+...

全部展开

1
2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²
=(2(x²+6x+1)+(x²+1))((x²+6x+1)+2(x²+1))
2
(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x+7)[(x^2+8x+7)+8]+15
=(x^2+8x+7)^2+8(x^2+8x+7)+15
=[(x^2+8x+7)+3][(x^2+8x+7)+5]
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)

收起

太简单了,先合并脱括号再分解就行了

1. 2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²
=[(x²+6x+1)+2(x²+1)]*[2(x²+6x+1)+(x²+1)]
=(3x²+6x+3)*(3x²+12x+3)
=3(x²+2x+1)*3(x...

全部展开

1. 2(x²+6x+1)²+5(x²+1)(x²+6x+1)+2(x²+1)²
=[(x²+6x+1)+2(x²+1)]*[2(x²+6x+1)+(x²+1)]
=(3x²+6x+3)*(3x²+12x+3)
=3(x²+2x+1)*3(x²+4x+1)
=3(x+1)²3*(x²+4x+1)
=9(x+1)²(x²+4x+1)
2. (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)+120
=(x²+8x+10)*(x²+8x+12)
=(x²+8x+10)*(x+2)*(x+6)
其实,这两道题就是障眼法,你只要把第一题的(x²+6x+1)看成是a,(x²+1)看成是b,那题目就变成了2a²+5ab+2b²,十字相乘一下子就出来了。第二题也是类似的想法,就不赘述了。

收起