x(1-1-x分之2)÷(x+1)-x²-2x+1分之x(x²-1)-搜答案-神马作文网

x^2-1=(x+1)(X-1)x^2-2x+1=(x-1)^2第一项为（X+1）/(x-1)第二项为-x(x+1)so最后-1/x(x-1)

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

（3-x分之x-1）平方÷（x平方-6x+9分之9-x平方）平方*x平方-2x+1分之6+2x=[(x-1)/(3-x)]²÷[(9-x²)/(x²-6x+9)]

{[(x+2)/(x^2-2x)]-[(x-1)/(x-2)]}/[(x+3)/x]={[(x+2)/(x^2-2x)]-[(x^2-x)/(x^2-2x)]}*[x/(x+3)]=[(-x^2+2x

你写的看不懂分子分母的代数式用括号括起来否则分不出来先通分【（（x-2）(x+2)-x(x-1)）/x(x+2)(x+2)】(x+2)/(x-4)=(x-4)/x(x+2)(x-4)=1/x(x+2)

(x+3)/(x²-1)÷[(x+3)(x-1)/(﹣x²+2x-1)]=(x+3)/(x²-1)÷[-(x+3)(x-1)/(x²-2x+1)]=(x+3)/

步步高点读机哪里不会点哪里

(x^2-2x分之x+2-x^2-4x+4分之x-1)/x分之4-x=[(x+2)/(x(x-2)-(x-1)(x-2)^2]/(4-x)/x=[(x+2)(x-2)-x(x-1)/x(x-2)^2]

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

x分之4再问：有过程吗他那个是x-1分之1-x+1分之1的再答：[1/(x-1)-1/(x+1)]/[x/（2x方-2）]=[2/（x方-1）]*[2（x方-1）/x]=（2*2）/x=x/4

(x-3)分之x-(x^2-3x)分之(x+6)+x分之1等于几?=x²/x(x-3)-(x+6)/x(x-3)+(x-3)/x(x-3)=(x²-x-6+x-3)/x(x-3)=

=[(x+2)/x(x-2)-(x-1)/(x-2)²]×x/(4-x)=[(x²-4-x²+4)/x(x-2)²]×x/(4-x)=1/(x-2)²

①X平方+2X分之X+2-X平方-4X+4分之X-1=(x+2)/x(x+2)-(x-1)/(x-2)²=[(x-2)²-x²+x]/x(x-2)²;=(4-3

你的表达有问题啊!请注意括号的正确使用,以免造成误解.［（x＋2）/（x^2－2x）－（x－1）/（x^2－4x＋4）］÷［（4－x）/x］＝｛（x＋2）/［x（x－2）］－（x－1）/（x－2）^2

(x²-1分之x²+1)-(x-1分之x-2)÷(x分之x-2)=(x+1)(x-1)分之(x²+1)-(x-1)分之(x-2)*(x-2)分之x=(x+1)(x-1)分

[（x+2)/x-(x-1)/(x-2)]÷（x-4)/x=[1+2/x-1-1/(x-2)]×x/(x-4)=(x-4)/x(x-2)×x/(x-4)=1/(x-2)=1/(3-2)=1这是我在静心

原式=[（x+2）/x（x-2）-（x-1）/（x-2）²]÷（4-x）/x=[（x+2）（x-2）/x（x-2）²-x（x-1）/x（x-2）²]÷（4-x）/x=[（

原式=[2x(x+1)/(x+1)(x-1)-x(x-1)/(x-1)²]÷x/(x+1)=[2x/(x-1)-x/(x-1)]×(x+1)/x=x/(x-1)×(x+1)/x=(x+1)/