(x+1)4+(x2-1)2+(x-1)4
来源:百度知道 编辑:UC知道 时间:2024/05/16 10:58:47
(x+1)4+(x2-1)2+(x-1)4
(x+1)^4+(x^2-1)^2+(x-1)^4
=(x+1)^4+(x+1)^2(x-1)^2+(x-1)^4
= (x+1)^4+2(x+1)^2(x-1)^2+(x-1)^4-(x+1)^2(x-1)^2
=(x +1+x-1)^2-(x+1)^2(x-1)^2
=(2x)^2-(x+1)^2(x-1)^2
=(2x+(x+1)(x-1))(2x-(x+1)(x-1))
=(x^2+2x-1)(x^2+2x+1)
=(x+1)^2(x^2+2x-1)
(x+1)^4+(x^2-1)^2+(x-1)^4
=(x^4+4x^3+6x^2+4x+1)+(x^4-2x^2+1)+(x^4-4x^3+6x^2-4x+1)
=3x^4+10x^2+3
因式分解的结果是
(3x^2+1)(x^2+3)
(x+1)4+(x2-1)2+(x-1)4=4x+4+4x-2+4x-4=12x-2
12x+2
(x2-1)(x2+x+1)/x(x2-3x-28)(x+2)<0
解方程x2+x+1=2/x2+x
(x+1)4+(x2-1)2+(x-1)4
分解因式: (1)x2-2x-1 (2)4(x-y+1)+y(y-2x) (3) (x2-2x)2-7(x2-2x)+12
|x2-3x-4|>x+1
(x-1)(x+1)(x2+x+1)(x2+x+1)
请给出以下不等式的过程:2x2-3x-2>2 -3x2+6x>0 4x2-4x+1>0 -x2+2x-3>0
设A={x/x2+4x=0} ,B=} x/x2+2(a=1)x+a2-1=0}
设集合A={x/x2+4x=0},集合B={x/x2+2(a+1)x+a2-1=0, a属于R}
若X1.X2都满足条件|2X-1|+|2X+3|=4,且X1<X2,则X1-X2的取值范围为______