(x² x分之x-1)除以x² 2x 1分之x² 2x 1分之-搜答案-大师作文网

原式=[(x+1)/x]÷[(2x²/2x)-(x²+1)/2x]=[(x+1)/x]÷[(2x²-x²-1)/2x]=[(x+1)/x]×[2x/(x&sup

x+2分之x平方-2x-x-1分之x平方-4X+4除以x平方-16分之x平方+4x=x(x-2)/(x+2)÷(x-2)^2/(x-1)÷[x(x+4)/(x+4)(x-4)]=[x(x-2)/(x+

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)/

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

①3除以x分之2=3*x/2=3x/2②x分之3除以x^2分之6=3/x*x^2/6=x/2③x分之2x+1除以5分之6+12x=(2x+1)/x*5/(6+12x)=5/6x④x^2+3x+2分之x

原式=[√(x/4)-2x√(1/x)]÷3√x=√(x/4)÷3√x-2x√(1/x)÷3√x=(1÷3)×√(x/4÷x)-(2x÷3)×√(1/x÷x)=1/3×√(1/4)-(2x/3)×√(

把平方代入根号除以根号得出｛3根号X-2根号X｝除以3根号X=根号X除以3根号X=3分之1

x^2-x-6分解因式得(x+2)(x-3)x^2-4分解因式得(x-2)(x+2)故(x^2-x-6分之x^2-4+x-3分之x+2)=(x-2)/(x-3)+(x+2)/(x-3)=2x/(x-3

原式=[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x+3)(x+5)]\(x^2+7x+13)=[(1分之x+2)-(1分之x+3)+(1分之x+

x的平方-2x+1分之x²-1+x-2分之2x-x²除以x=（x+1）（x-1）/（x-1）²-x(x-2)/(x-2)÷x=（x+1）/(x-1)-1=(x+1-x+1

x^2-1分之x^2-3x+2除以（x+1分之2x-1-x+1)=(x+1)(x-1)分之(x-1)(x-2)除以【x+1分之2x-1-（x+1）分之x²-1】=(x+1)分之(x-2)÷【

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

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

直接带入算再问：必须要过程的，再答：见图

X²+x-2分之1+x除以x+2分之x²-1,=（x+1）/(x+2)(x-1)÷（x+1）（x-1）/(x+2)=1/(x-1)²=1/(1/2-1)²=4

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

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答：[x/(x-1)+1/(x-1)]÷[(x+1)/(x²-2x+1)]=[(x+1)/(x-1)]÷[(x+1)/(x-1)²]=[(x+1)/(x-1)]×[(x-1)