(x-1分之3-x-1)÷x的平方-2x 1分之x-2其中x是不等式组

[(3-x)/(x-1)-1]/[(x-2)/(x^2-2x+1)]=[(3-x-x+1)/(x-1)]*[(x^2-2x+1)/(x-2)}=[(-2x+4)/(x-1)]*[(x^2-2x+1)/

{[(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

1:x²+X-2分之3=X-1分之X-X+2分之XX²+X-(X-1分之X)+X-2分之X=0X²+2X-(X-1分之X)-2分之X=0∵X-1≠0所以（X-1)(X&#

只有我能看懂吧?!再答： 再答：题是这样的吗？再答： 再问：你终于回来了再问：在吗再答：在再问：等等我去写题再问：去吧

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

x/(x^2+x+1)=-1/2即-2x=x^2+x+1x^2+3x+1=0(x-1)/(2x^2+x^3-1)/(x^2-x+1)=(x-1)/(2x^2+x^3-1)(x^2-x+1)=(x-1)

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

=x[1/(x-1)-1/(x+1)]×[-(x-1)/2x]=x(x+1-x-1)/(x-1)(x+1)×[-(x-1)/2x]=-1/(x+1)当x=√3时原式=-1/(√3+1)=-(√3-1)

1题目是：x²/(x²-2)=3,求[1/(1-x)-1/(1+x)]/[x/(x²-1)+x]吧?如果是这样,那么由x²/(x²-2)=3得,x&#

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

(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-1)(x-3)(x-4)(x-5)-(x-1)(x-2)(x-3)(x-5)=(x-1)(x-2)(x-4)(x-5)-(x-1)(x-2)(x-3)(x-4)(x-1)(x-3)(x-5)(x

“数理答疑团”为您解答,希望对你有所帮助.x²+x-2=0x=-2和x=1是分式方程x的平方+x-2分之3=x-1分之x-x+2分之x的增根手机提问的朋友在客户端右上角评价点【满意】即可.

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你的公式写出来没办法断句,比如说,“x-3分之x的平方”,是说分子为x的平方,分母为x-3；还是说分子为x,分母为x-3,然后整体平方；还是说是三分之一倍的x平方,再整体被3减?能不能传图上来?

x/(x-3)-(x+6)/(x^2-3x)+1/x=[x^2-(x+6)+(x-3)]/x(x-3)=(x^2-9)/x(x-3)=(x+3)/x.

两边乘以x(x-1)得3(x-1)+6x=73x-3+6x=79x=10x=10/9经检验：x=10/9是方程的解

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)/