(2x-3分之x-1)除以x²-9分之x-搜答案-灵鹊做题网

x≠-2,-3,-4

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

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^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)(

X^5分之1-X吧（1-x^2）=(1-X)(1+X)（x^3+x^2）=x^2（X+1）约分等于x^2分之(1-X)再乘x^2分之1乘x分之1

两边乘(x-1)(x+2)x²+2x-x²-x+2=3x=1经检验,x=1时分母x-1=0增根,舍去方程无解

[(x+3)/(x-3)]/[(x+2)/(x-4)]=[(x+3)(x+2)]/[(x-3)(x-4)](x-3)(x-4)≠0即x≠3且x≠4

x²-2分之x²=33x²-6=x²2x²=6x²=3(1-x分之1-1+x分之1)除以（x²-1分之x+x)=[(1-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)

您好：[(X^2-2X)分之(X+2)-(X^2-4X+4)分之(X-1)]除以x-4/x^3-2x^2=[(x+2)/x(x-2)-(x-1)/(x-2)²]*x²(x-2)/(

（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)

原式=(x^2-8x+16)^3(x+1)^6/(x^2-2x+1)^3(x-4)^4(x+2)^2=(x-4)^6(x+1)^6/(x-1)^6(x-4)^4(x+2)^2=(x-4)^2(x+1)

答：[(x-2)/(x-1)]÷[x+1-3/(x-1)]=[(x-2)/(x-1)]÷[(x²-1-3)/(x-1)]=[(x-2)/(x-1)]×(x-1)/(x²-4)=1/