先化简再求值1-x的平方分之(1 x)-搜答案-大师作文网

8-2x分之x的平方-16=[(x+4)(x-4)]/[(-2)(x-4)]=-(x+4)/2把x=-2代入上式得：－1

是(x^2/(x-1)-x+1)/(4x^2/(1-x)-4x+1)吗原式=(x-x+1)(x+x-1)/(x-1)/((4x^2-4x+1+4x^2-x)/(1-x))=-(2x-1)/(4x^2-

{（x的三次方-x）分之（x的平方-2x+1）}*（1-x分之1）=【(x-1)^2/[x(x+1)(x-1)]】*(x-1)/x=[(x-1)/[x(x+1)]]*(x-1)/x=(x-1)^2/[

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

(x分之X的平方+4-4)除的X平方+2x分之X的平方-4=[(x²+4)/x-4]÷[(x²-4)/(x²+2x)]=[(x²-4x+4)/x]÷[(x+2)

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=-1/2∴2x=2×(-1/2)=-1

原式=[3x/(x-1)-x/(x+1)]×[(x²-1)/x]={3x(x+1)/[(x-1)(x+1)]-x(x-1)/[(x-1)(x+1)]}×[(x-1)(x+1)/x]=[3x(

（x-1分之1-x+1分之1）/2x的平方-2分之x=[(x+1-x+1)/(x-1)(x+1)]/2x²-x/2=1/x²(x-1)(x+1)-x/2x=2=1/[2²

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

/>（2x-x分之1+x的平方）除以x分之x-1,其中x=根号2=[2x²-(1+x²)]/x÷(x-1)/x=(x²-1)/(x-1)=(x-1)(x+1)/(x-1)

分子分母同乘以4.再一化简

（x+1）÷[2+x分之(1+x²)]=（x+1）÷[x分之(2x+1+x²)]=(x+1)÷[x分之(x+1)²]=(x+1)·[(x+1)²分之x]=(x+

x的平方+3X分之X的平方+8X+X的平方+6X+9分之X的平方-9=x(x+3)分之x(x+8)+(x+3)²分之(x+3)(x-3)=(x+3)分之(x+8)+(x+3)分之(x-3)=

x²+x-2=0(x-1)(x+2)=0x=1或x=-2(x²-x+1)/(x-1)÷(4x²-4x+1)/(1-x)=(x²-x+1)/(x-1)*(1-x)

x+1/2=0,y-1=0x=-1/2,y=1原式=4x²y-(6xy-8xy+4-x²y)+1=4x²y-(-2xy+4-x²y)+1=4x²y+2