(x -y分之1-x y分之1)除以x的平方-2xy y的平方分之2y

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通分x-y分之1与x²-y²分之xy+y²

1/(x-y)=(x+y)/(x²-y²)(xy+y²)/(x²-y²)=(xy+y²)/(x²-y²)

(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

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

计算(3y分之2x)×(4y分之3x)÷(4分之1xy)

=(2X/3Y)*(3X/4Y)*(4/xy)=(2x*3x*4)/(3y*4y*xy)=2x/(y^3)

先化简,再求值:x的平方-y的平方分之x的平方+2xy+y的平方除一x-y分之x的平方+xy,其中x=√2-1,y=√2

x的平方-y的平方分之x的平方+2xy+y的平方除一x-y分之x的平方+xy,其中x=√2-1,y=√2+1=(x+y)^2/(x+y)(x-y)/((x^2+xy)/(x-y))=(x+y)/(x-

y分之x等于2分之1 xy怎么算

绝对算不出来再答:如果是一道题发图咯再问:x平方减xy加y平方分之x平方加2xy加y平方他给你条件是y分之x等于2分之1xy再问:吧xy无视掉再答:马上再答:化成Y等于2X在带入再问:然后怎么样再答:

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:根号下X(X+Y)除根号下X+Y分之XY方

是否这样:√[x+y)/xy²]÷√x(x+y)=√[(x+y)/(x+y)x²y²]=√1/x²y²=1/|x||y|再问:是这样麻烦了~再答:知道

(x+2y-3xy+2x+y-xy)-(-2-y+xy),其中x+y=2分之1,xy=-2分之1

(x+2y-3xy)-(-2x-y+xy)=x+2y-3xy+2x+y-xy=(1+2)x+(2+1)y-(3+1)xy=3x+3y-4xy=3(x+y)-4xy=3*1/2-4*(-1/2)=3/2

(-3分之1xy)的平方×[xy(2x-y)-2x(xy-y的平方)]

原式(-xy/3)²*[xy(2x-y)-2x(xy-y²)]=(x²y²/9)(2x²y-xy²-2x²y+2xy²)

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

已知x-y分之xy= - 3分之1,求x-2xy-y分之2x+3xy-2y的值

x-y分之xy=-3分之1xy分之(x-y)=-3x-y=-3xyx-2xy-y分之2x+3xy-2y=(-6xy+3xy)/(-3xy-2xy)=3/5

已知x分之1减y分之1等于4,求x+7xy-y分之x-2xy-y

由已知条件可得:x-y=-4xy,(也就是在等式两边乘xy后整理,即可),代入所求的式子里分母为-4xy+7xy,即3xy;分子为-4xy-2xy,即-6xy,化简得-2

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)/

(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-1分之y= 分之xy-y

分子上应写:(x-1)²或者x²-2x+1再问:怎么书面表达呢?能否详细一些,谢谢再答:希望你及时采纳!!!!!