{x y=18,30% 75%=18×50%用二元一次方程解下列实际问题

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已知XY为正数,X+Y=1 求1/XY+XY的最小值

令F(XY)=1/XY+XY,当XY=1的时候,F(XY)=2,最小.(可由函数图形象得出).XY趋于正无穷大的时候F(XY)趋于正无穷大,XY无限趋于零的时候F(XY)趋于正无穷大.所以XY越接近1

先化简,再求值 ⒈2(Xy+Xy)-3(Xy-xy)-4Xy,其中X=1,y=-1

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

[(xy+2)(xy-2)-2x^2y^2+4]/xy,其中x=10,y=-1/25

[(xy+2)(xy-2)-2x2y2+4]÷(xy)=(x2y2-4-2x2y2+4)÷xy=-x2y2÷xy=-xy,当x=10,y=1/25时,原式=-2/5.

[(xy+2)(xy-2)-2x^2y^2+4]/(xy),其中:x=10,y=-1/25

[(xy+2)(xy-2)-2x^2y^2+4]/(xy)=[(x^2y^2-4)-2x^2y^2+4]/(xy)=-x^2y^2/xy=-xy=-10*(-1/25)=2/5

先化简,再求值:{(xy-2)^2+4xy-4}/(-xy),其中x=10.,y=-1/25

令t=xy方程变为:[(t-2)^2+4t-4]/(-t)=(t^2-4t+4+4t-4)/(-t)=-t=-xy又x=10,y=-1/25-xy=-10*(-1/25)=2/5

化简求值:[(xy-2)²+4xy-4]÷(xy),其中x=10,y=1/25

[(xy-2)²+4xy-4]÷(xy)=(x²y²-4xy+4+4xy-4)÷(xy)=x²y²÷xy=xy=10×1/25=2/5

xy'=y+xy的

xdy=(y+xy)dxdy/y=((1+x)/x)dxln|y|=ln|x|+x+cy=±e^(ln|x|+x+c)其中c是常数再问:真还不理解我们是选择题:y=cxe^xy=c+x-x^2y=cs

已知(x+1)²+|y-1|=0,求2(xy-5xy²)-(3xy²-xy)得值

解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(

当x=3,y=3分之1时,求代数出3xy-[2xy-2(xy-2分之3xy)+xy]+3xy的值

3xy-[2xy-2(xy-2分之3xy)+xy]+3xy=6xy-[2xy-2xy+3xy+xy)=6xy-4xy=2xy=2×3×3分之1=2

xy+e^(xy)=1,求y的导数

该题为隐函数求导.xy+e^(xy)=1则y+xy'+e^(xy)(y+xy')=0解得:y'=-y/x解答完毕.

若x>1,y>0且满足xy=xy,xy=x

由题设可知y=xy-1,∴x=yx3y=x4y-1,∴4y-1=1,故y=12,∴12x=x,解得x=4,于是x+y=4+12=92.故答案为:92.

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

当x=3,y=-1/3时求代数式3x*xy-[2xy*y-2(xy-2/3x*xy)+xy]+3xy*y的值

=3x^2y-2xy^2+2(xy-3/2x^2y)-xy+3xy^2=3x^2y+xy^2+2xy-3x^2y-xy=xy+xy^2=3*(-1/3)+3*1/9=-2/3

已知xx+xy=1,xy-yy=-4,则xx+2xy-yy=

xx+2xy-yy=-3两式相加即可

(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x+y=1,xy=-3求多项式(x+xy)-[(xy-2y)-x]-(-xy)的值在线等,(8:30之前谁告我,我就

(x+xy)-[(xy-2y)-x]-(-xy)=(x+xy)-(xy-2y-x)+xy=x+xy-xy+2y+x+xy=2(x+y)+xy=2*1-3=-1设这个多项式为f(x).则f(x)+2(x

xy+x=20 xy+y=18

由题意得:X=Y+2.那么Y(Y+2)+Y+2=20(Y+2)×(Y+1)=20所以y=3那么x=5可待入xy+y=18就不对了.(Y+2)×(Y+1)=20,Y应该是-6,X是-4,答案就对了.X=

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy),其中x=20,y=-25分之1

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+