已知一个分式乘以2x^3 y
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1/x+1/y=3∴x+y=3xy3x-2xy+3y/x+xy+y=[3(x+y)-2xy]/[(x+y)+xy]=(9xy-2xy)/(3xy+xy)=7/4
2x-3y=0所以y=2x/3代入(x^2+xy-y^2)/(2x^2+xy-y^2)得到[x^2+x*2x/3-(2x/3)^2]/[2x^2+x*2x/3-(2x/3)^2]=(x^2+2x^2/
1)由1/x-1/y=5,得(y-x)/xy=5,所以y-x=5xy,即x-y=-5xy(3x+5xy-3y)/(x-3xy-y)=[3(x-y)+5xy]/[(x-y)-3xy]=(-15xy+5x
∵4x-y=0∴y=4x∴(x²+3y)/(2y-3x)=(x²+12x)/(8x-3x)=[x(x+12)]/(5x)=(x+12)/5明白请采纳,有疑问请追问!有新问题请求助,
2X-3Y=02X=3YX²+Y²/X²-3XY=(4X^2+4Y^2)/(4X^2-12XY)=(9Y^2+4Y^2)/(9Y^2-18y^2)=13y^2/-9y^2
∵3x-5y=0,∴x=5y3,∴原式=5y3−2y5y3+3y=-111.
首先化简通分1/x+1/y=5得:(x+y)/xy=5即有x+y=5xy将要求的分式化简:3x+xy+3y/x-2xy+y=3x+3y+xy/x+y-2xy(根据加法交换律)=3(x+y)+xy/(x
|y/x|=3,=>y=3x或者y=-3x,x不等于01:y=3x;[(x^2+2xy+y^2)/(x^2*y^2)]*[(x-y)/(x-2y)]=16/25;x,y不等于02:y=-3x;[(x^
=-(x+y)-3/5=-5-3/5=-5.6
x/2=y/3x=2y/3(3x+2y)/(2x-3y)=[3(2y/3)+2y]/[2(2y/3)-3y]=4y/(-5y/3)=-12/5
因为:1/x-1/y=3(y-x)/xy=3x-y=-3xy(1)又因为:(2x+3xy-2y)/(x-2xy-y)=[2(x-y)+3xy]/[(x-y)-2xy]代入(1)得:(-6xy+3xy)
1/x-1/y=3(y-x)/xy=3y-x=3xy(4x-5xy-4y)/(3x-2xy-3y)=(4(x-y)-5xy)/(3(x-y)-2xy)=(4*-3xy-5xy)/(3*-3xy-2xy
x²+3分之x-2y的值为0∵x²+3>0∴x-2y=0∴x=2yx²-y²分之x²-3xy-4y²=(x-4y)(x+y)/(x+y)(x
因为已知x/y=2,所以x=2y代入算式得:(4y*y-2y*y+3y*y)/(4y*y+2y*y+6y*y)化简原式=5/12
分子分母同除以y²,看我上传的图片.
y-x=3xyx-y=-3xyx-2xy-y=-5xy2x+3xy-2y=2(x+y)+3xy=-6xy+3xy=-3xy分式x-2xy-y分之2x+3xy-2y=5/3
即xy=2²-(√3)=4-3=1原式=[(x+y)²-(x-y)²]/(x+y)(x-y)*[-(x²-y²)/x²y²]=(x
3x-2y/2x-3y=[3(x/y)-2]/[2(x/y)-3]分子分母同时除以y=(4-2)/(8/3-3)=2/(-1/3)=-6
∵x-y/xy=3∴x-y=3xy2x+3xy-2y/x-2xy-y=[2(x-y)+3xy]/[(x-y)-2xy]=(6xy+3xy)/(3xy-2xy)=9xy/(xy)=9