怎样把【x(1+x’*x‘)-x’(1+x*x)】/(1+x*x)(1+x‘*x’)因式分解,*为乘号,x与x‘是2个未
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/18 02:38:40
怎样把【x(1+x’*x‘)-x’(1+x*x)】/(1+x*x)(1+x‘*x’)因式分解,*为乘号,x与x‘是2个未知数
![怎样把【x(1+x’*x‘)-x’(1+x*x)】/(1+x*x)(1+x‘*x’)因式分解,*为乘号,x与x‘是2个未](/uploads/image/z/1406712-48-2.jpg?t=%E6%80%8E%E6%A0%B7%E6%8A%8A%E3%80%90x%EF%BC%881%2Bx%E2%80%99%2Ax%E2%80%98%EF%BC%89-x%E2%80%99%EF%BC%881%2Bx%2Ax%EF%BC%89%E3%80%91%2F%EF%BC%881%2Bx%2Ax%EF%BC%89%EF%BC%881%2Bx%E2%80%98%2Ax%E2%80%99%EF%BC%89%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%2C%2A%E4%B8%BA%E4%B9%98%E5%8F%B7%2Cx%E4%B8%8Ex%E2%80%98%E6%98%AF2%E4%B8%AA%E6%9C%AA)
把x*写成y得:
原式=(x(1+y²)-y(1+x²))/(1+x²)(1+y²)
=(x+xy²-y-x²y)/(1+x²)(1+y²)
=(x-y)(1-xy)/(1+x²)(1+y²)
原式=(x(1+y²)-y(1+x²))/(1+x²)(1+y²)
=(x+xy²-y-x²y)/(1+x²)(1+y²)
=(x-y)(1-xy)/(1+x²)(1+y²)
怎样把【x(1+x’*x‘)-x’(1+x*x)】/(1+x*x)(1+x‘*x’)因式分解,*为乘号,x与x‘是2个未
化简:(1)2/3x*根号9x+6x*根号x/4(x>0) (*为乘号,
[x(x+1)+x平方(x-1)]+(-3x)
(x+1)(x+2)(x+3)(x+6)+x的平方因式分解
(x/x-1)+(5x-2)/x*x-x=1 求x(有图)
因式分解x^4-x^-2-x+1
因式分解:x(x-1)(x-2)-6
(x-1)x(x+1)-24因式分解,中间的那个是字母x,不是乘号
若1+x+x*x+x*x*x=0,求x+x*x+x*x*x+.+x(2000次幂)
(x-1)(x-3)-8 因式分解
因式分解10x^2+5x是等于5x(2x+1)吗
因式分解:(x²-x)²-(x-1)²