x²-2x 1分之x除以(x²-1分之x 1 1),其中x=根号2-1-搜答案-搜搜考试网

y=[x/(x-2)-x/(x+2)/2x/(2-x)=4x/(x-2)(x+2)*(2-x)/2x=-2/(x+2)y的值是整数,x+2=1,-1,2,-2x=,-1,-3,0,-4

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

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

x1+x2=4x1x2=1/2原式=(x1+x2)²÷(x1+x2)/x1x2=x1x2(x1+x2)=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)(

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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)

4-4x+x^2分之2x-6除以（x+3)乘以3-x分之x^2+x-6=2(x-3)/(x-2)²×1/(x+3)×(x+3)(x-2)/(3-x)=-2/(x-2)

2x分之(x+2)*(x^2+5x+6)分之(x^3-3x^2)÷(x^2+2x-3)分之(x^3-9x)=2x分之(x+2)*(x+2)(x+3)分之x^2(x-3)*x(x+3)(x-3)分之(x

（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)

（x2-4x+x2/4-4+x+(2-x)/2）/（x/x-2）=(5/4x2-3x-4+(2-x)/2）/（x/x-2）=1/4(5x2-12x-16+4-2x)/（x/x-2)=（5x2-14x-

x1+x2=4x1x2=-1(x1+x2)^2/(1/x1+1/x2)=(x1+x2)^2*x1x2/(x1+x2)=x1x2*(x1+x2)=-4

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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)