cos4x怎么化简 化简分式x-（1/1-x）-（x^3-3x+1/x^2-1）,并求x=-1/3时该分式的值。

化简分式x-（1/1-x）-（x^3-3x+1/x^2-1）,并求x=-1/3时该分式的值。  

化简分式x-（1/1-x）-（x^3-3x+1/x^2-1）,并求x=-1/3时该分式的值。

x-（1/1-x）-（x^3-3x+1/x^2-1）
＝x-(1/1-x)-{x(x^2-1)-2x+1}/x^2-1
=x+(1/x-1)-x+(2x-1)/x^2-1
=(2x-1)/x^2-1+(x+1)/(x+1)(x-1)
=(2x-1+x+1)/x^2-1
=3x/x^2-1
=(3x-1/3 )/(-1/3)^2-1
=-1/(-8/9)
=9/8

化简分式： 3x-x^3 -2 / (1-x)(1-x^3)

(3x-x^3 -2) / (1-x)(1-x^3)
=-(x-1)^2(x+2)/(x-1)^2(x^2+x+1)
=-(x+2)/(x^2+x+1)

x-[(1/1-x)+(x^3-3x+1/x^2-1)/(a-x^2)]

=x-[-1/(x-1)+(x³-3x+1)/(x+1)(x-1)]/[-(x²-4)]
=x+[(-x-1+x³-3x+1)/(x+1)(x-1)]/(x²-4)
=x+[(x³-4x)/(x+1)(x-1)]/(x²-4)
=x+[x(x²-4)/(x+1)(x-1)]/(x²-4)
=x+x/(x+1)(x-1)
=(x³-x+x)/(x+1)(x-1)
=x³/(x²-1)

化简分式：1/(x²+3x+2)+1/(x²+5x+6)+1/(x²+7x+12)

原式=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/(x+1)-1/(x+4)
=3/(x²+5x+4)

1.当分式3x^2-6x+3/(1-x)^3的值是正整数时，求分式（x/x^2-x-2）+1/x-

(3x^2-6x+3)/(1-x)^3的值是正整数
(3x^2-6x+3)/(1-x)^3 = 3(1-x)^2/(1-x)^3 = 3/(1-x)为正整数
1-x=3，或1
x=-2，或0
x/(x^2-x-2) + 1/(x-2)
= {x+(x-1)}/{(x+1)(x-2)}
= (2x+1)/{(x+1)(x-2)}
当x=-2时：= (-4+1)/{(-2+1)(-2-2)} = -3/4
当x=0时：= (-1)/{(0+1)(0-2)} = 1/2

1/(x平方+3x+2)+1/(x平方+5x+6)+1/(x平方+7x+12)化简分式

1/(x平方+3x+2)+1/(x平方+5x+6)+1/(x平方+7x+12)
=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/(x+1)-1/(x+4)
=3/(x+1)(x+4)

初一数学：化简分式 1/X+1/(X+a)+1/(x+2a)+1/(x+3a)+1/(x+4a)

设：x+2a=t；则x=t-2a，x+a=t-a，x+3a=t+a，x+4a=t+2a。所求为：1/(t-2a)+1/(t-a)+1/t+1/(t+a)+1/(t+2a)=2t/(t²-4a²)+2t/(t²-a²)+1/t=(5倍t的四次方-15a²t²+4倍a的四次方)/(t*(t²-4a²)*(t²-a²))。最后把t用x+2a换回去。完毕！

化简分式：[（3x/x－1）－（x/x+1）]÷x/x²－1

[（3x/x－1）－（x/x+1）]÷x/x^2－1
=[(3x(x+1)-x(x-1))/(x-1)(x+1)]÷x/(x^2-1)
=(3x^2+3x-x^2+x)/(x^2-1) × (x^2-1)/x
=(2x^2+4x)/(x^2-1) × (x^2-1)/x
=x(2x+4)/(x^2-1) × (x^2-1)/x
=2x+4

化简分式：2/x-1 ÷ (2/x²-1 + 1/x+1）

原式=(2/x-1) ÷(2+x-1)
=(2/x-1) *(1/x+1)
=2/x²-1

先化简分式(x+3)/(x^2-1)÷(x^2+6x+9)/(x^2-2x+1)+1/(x+1),再去适当的x的值代入取

解：
原式=[(x+3)/(x²-1)]÷[(x²+6x+9)/(x²-2x+1)]+1/(x+1)
={(x+3)/[(x-1)(x+1)]÷[(x+3)²/(x-1)²]+1/(x+1)
={(x+3)/[(x-1)(x+1)]}×[(x-1)²/(x+3)²]+1/(x+1)
=(x-1)/[(x+1)(x+3)]+1/(x+1)
=(x-1)/[(x+1)(x+3)]+(x+3)/[(x+1)(x+3)]
=(x-1+x+3)/[(x+1)(x+3)]
=(2x+2)/[(x+1)(x+3)]
=2(x+1)/[(x+1)(x+3)]
=2/(x+3)
当x=2时
原式=2/(2+3)=2/5