化简y-x分之1的差 x-y分之1的差

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(X+Y分之X+X+Y分之2Y)乘X+2Y分之XY除以(X分之1+Y分之1)要有过程

原式=[(x+2y)/(x+y)][xy/(x+2y)]÷[(x+y)/xy]=[xy/(x+y)]×xy/(x+y)=x²y²/(x+y)²

画出y=x分之1(x

这是反比例函数.图像是位于一、三象限的双曲线.由于x<0,因此函数图象为双曲线的第三象限部分手工绘图应用描点法.见下图

x的平方-y的平方分之2xy + x+y分之x - y-x分之y

(x的平方-y的平方)分之2xy+(x+y)分之x-(y-x)分之y=(x的平方-y的平方)分之2xy+(x+y)分之x+(x-y)分之y=(x²-y²)分之[2xy+x(x-y)

已知x分之1-y分之1=2001,求分式x-y分之x-xy-y的值.

1/x-1/y=2001则(x-xy-y)/(x-y)=2002/2001

(x分之x-y)-(y分之x+y)+(2xy分之x的平方-y的平方)

(x分之x-y)-(y分之x+y)+(2xy分之x的平方-y的平方)=1-y/x-1-x/y+(x^2-y^2)/2xy=-2(y^2+x^2)/2xy+(x^2-y^2)/2xy=-(x^2+3y^

1 x^2y分之x^4

1x^2y分之x^4=x²/y2ab分之ab-b^2(b不等于0)=1-b/a36abc^2分之36ab^3c=6b²/c4a^2-b^2分之(a+b)^3=(a+b)²

已知x分之1+y分之1=-2,则x+xy+y分之x-xy+y的值等于

1/x+1/y=-2则(x-xy+y)/(x+xy+y)分子分母同除以xy=(1/x+1/y-1)/(1/x+1/y+1)=(-2-1)/(-2+1)=3

计算:1、(xy分之x²-y²)×(x-y分之x)-(y分之x);

1、(xy分之x²-y²)×(x-y分之x)-(y分之x);=(x+y)(x-y)/xy×x/(x-y)-x/y=(x+y)/y-x/y=(x+y-x)/y=12、6xy²

(x+y分之x加x+y分之2y)乘x+2y分之xy除(x分之1加y分之1)等于?

原式=[(x+2y)/(x+y)]×[xy/(x+2y)]÷[(x+y)/xy]=[(x+2y)/(x+y)]×[xy/(x+2y)]×[xy/(x+y)]=x²y²(x+2y)/

xy分之x的平方-y的平方÷〈1分之x-y-1分之x+y〉

原式=xy/(x-y)(x+y)÷(2y)/(x-y)(x+y)=x/2再问:再问:

x-y分之1-x+y分之1)除以x的平方-y的平方分之xy的平方

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

x-y分之1+x+y分之1)除以x的平方-y的平方分之xy

现将括号里的通分x2-y2可约去最后的y分之2再问:能具体一点吗再答:原式=((x+y+x-y)/(x2-y2)÷(x的平方-y的平方分之xy)=2x/(x2-y2)×(x2-y2)/xy=y分之2只

1,x分之x+y减去x-y分之x加上x的平分-xy分之y的平分

解1题原式=[(x+y)/x]-[x/(x-y)]+[y²/(x²-xy)]=[(x+y)/x]-[x/(x-y)]+{y²/[x(x-y)]}={(x+y)(x-y)/

数学题(1-y+x分之y)÷y的平方-x的平方分之x

(1-y+x分之y)÷y的平方-x的平方分之x=[(y+x-y)/(y+x)]÷x/(y²-x²)=x/(y+x)*[(y+x)(y-x)/x]=y-x

(x+y分之x + x+y分之2y)乘x+2y分之xy 除以(x分之1+y分之1)同样用两...

是这样的形式吗?[x/(x+y)+2y/(x+y)]×[xy/(x+y)]÷(1/x+1/y)一:原式=(x+2y)/(x+y)×[xy/(x+2y)]÷[(x+y)/xy]=x²y&sup

x的平方-64y的平方分之2x-x-8y分之1

x的平方-64y的平方分之2x-x-8y分之1=[2x-(x+8y)]/(x-8y)(x+8y)=(x-8y)/(x-8y)(x+8y)=1/(x+8y)

计算:(1)x-y分之2x的平方-y-x分之x的平方-4xy+x-y分之2y的平方-x的平方

x-y分之2x的平方-y-x分之x的平方-4xy+x-y分之2y的平方-x的平方=2x/(x-y)-(x²-4xy)/(y-x)+(2y-x²)/(x-y)=(2x+x²

y=1-x的平方分之sinx,求y'.

y’=[sinx/(1-x²)]'=[(sinx)'(1-x²)-sinx(1-x²)']/(1-x²)²=[cosx(1-x²)+2xsi