规定x y=xy分之Ax y,且5 6=6 5,
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x-4√xy-5y=0(√x+√y)(√x-4√y)=0√x=4√yx=16y(x+√xy-y)/(2x+√xy+3y)=(16y+4y-y)/(32y+4y+3y)=19y/39y=19/39
已知x=(√3-√2)/(√3+√2),xy=1,则可知x与y是倒数关系.即y=(√3+√2)/(√3-√2)代入3x的平方+5xy+3y的平方=3*(x+y)的平方-xy=3*{(√3-√2)的平方
x-3y=0且xy≠0,所以x=3y(x≠0,y≠0)带入分式,(9y-3y+y)/(9y+3y+y)=7/13
x+6y=5√xy(√x-2√y)(√x-3√y)=0√x=2√y√x=3√y所以x=4y或x=9y√x=2√y原式=(8y+2y-y)/(4y+2y+2y)=9/8√x=3√y原式=(18y+3y-
x+6y=5根号xyx+6y-5根号xy=0(√x-2√y)(√x-3√y)=0√x-2√y=0或√x-3√y=0(1)√x-2√y=0√x=2√yx=4yx+根号xy+2y分之2x+根号(xy-y)
x-5xy+6y=0(√x-2√y)(√x-3√y)=0所以√x=2√y,√x=3√yx=4y或x=9y原式=(8y)/(9y)=8/9或=(14y)/(20y)=7/10
3△5=(3+5)/(2*3*5)=8/(2*15)=4/15
可以先设xy=t原式=[4(t-1)^2-(t+2)(2-t)]/(1/4*t)=4(4t^2-8t+4-2t+t^2-4+2t)/t=4(5t^2-8t)/t=(20t-32)/t=20-32/t将
xy/(x+y)=2∴xy=2(x+y)(3xy-x-y)/(3x-5xy+3y)=[6(x+y)-x-y]/[3x+3y-10(x+y)]=5(x+y)/[-7(x+y)]=-5/7
因为X+Y分之XY=2,所以XY=2(X+Y)代入后面的分子式得:13(X+Y)/5(X+Y)=13/5
答案:第一空a=1第二空12/11
B=4,a=正负3次根号5
xy-y^2=y(x-y)=5因为x和y都是正整数,所以y=1y=52/5(xy^3-y^4)=2/5y^2(xy-y^2)=2y^2当y=1时,原式=2当y=5时,原式=50
(3x^2+2xy)/xy-(2x^2-xy)/xy=(3x^2+2xy-2x^2+xy)/xy=(x^2+3xy)/xy=x(x+3y)/xy=(x+3y)/y
[4(xy-1)^2-(xy+2)(2-xy)]除以4分之1xy=[4x^2乘以y^2-8xy+4-(4-x^2y^2)]除以4分之1xy=(4x^2乘以y^2-8xy+4-4+x^2y^2)除以4分
2分之1或负4分之3
因为x>0,y>0,所以将x-根号xy-2y=0因式分解得(根号x-2根号y)(根号x+根号y)=0所以根号x=2根号y,即x=4y后面请自己代入吧,我看不懂你的表述.不好意思了
5xy^2+axy^b=-2xy^b5xy^2=(-2-a)xy^b对比方程左右两边:首先知道方程左边不恒为0,因此两边的未知数幂的次数有各自对应关系.常数项对应常数项.即:2=b5=-2-aa=-7
1.x=1,y=1,√xy+y分之x+√xy=12.将△DFC绕点D顺时针旋转180°,得到△DGB,连接EG.则△DFC≌△DGB.所以FD=DG,BG=FC,角FDC=角BDG有因为DE.DF分别
∵x-y=4xy∴原式=[2(x-y)+3xy]/[-(x-y)-2xy]=(8xy+3xy)/(-4xy-2xy)=11xy/(-6xy)=-11/6