y= x y=0.25x²,y=1三条曲线围成图形的面积

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x>1,y>0,且满足xy=x^y,x/y=x^3y,求x+y

x=4,y=0.5,x+y=4.5(与人家的做法一样……)(1)解题思路是以S3为基准,用S3表示出S1,S2,S4即可.在三角形BCD中有:S2/S3=DF/CF,故S2=(DF/CF)S3;同理,

微分方程y'=xy+x+y+1的通解是?

dy/dx=xy+x+y+1dy/dx=(x+1)(y+1)分离变量dy/(y+1)=dx*(x+1)两边积分ln(y+1)=(x²/2)+x+lnC两边取以e为底的幂y+1=Ce^[(x&

已知:x-y=1,xy=-2.求:(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)的值

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15

已知x+y=-1,xy=-2,求代数式-5(x+y)+(x-y)+x(xy+y)的值

答:x+y=-1,xy=-2-5(x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)

1\x+1\y=8\x+y,则y\xy+x\xy=?

你确定没写错题目?后面的式子一约就变成前面的式子了,答案也是8/xy再问:怎么个约法?再答:y/xyyy约掉剩下1/x

已知:xy+x=-1,xy-y=-2.

(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

已知3/(x-y)=1/xy 求(-x-2xy+y)/ (2x+3xy-2y)

3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9

若x-y=4,xy=1,求(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)的值

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*

已知x+y=-1,xy=-2,求代数式-5(x+y)+(x-y)+2(xy+y)的值

-5(x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4(x+y)+2xy=-4×(-1)+2×(-2)=4+(-4)=0你有问题也可以在这里向我提问

(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

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

(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

已知X+Y=-1,xy=-2,求代数式-5(x+y)+(x-y)+(xy+y)的值

这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-

已知xy>0,证明xy+xy/1+x/y+y/x>=4

xy+1/xy>=2√(xy*1/xy)=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√(x/y*y/x)=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求

已知xy/x+y=1/2,则代数式3x-5xy+3y/-x+3xy-y=

因为xy/(x+y)=1/2所以x+y=2xy原式=3(x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

x²+xy/(x-4y)(x+y),其中x=4,y=-1

x²+xy/(x-4y)(x+y)=x(x+y)/(x-4y)(x+y)=x/(x-4y)=4/(4+4)=1/2

解方程x*x+xy=y+y*y

令y=kxx*x+kx*x=k*x+k*k*x*x(1-k*k+k)x^2-kx=0x((1-k*k+k)x-k)=0由上式得X=0或(1-k*k+k)x-k=0解得:k=(x-1+(或-)√((1-

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!!!!!

.已知:x-y=1,xy=-2.求:(-2xy+2 x+3y)-(3xy+2y-2x)-(x+4y+xy)的值

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采