x-xy=20,xy-y=12则x-y=?
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x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17
[2x(x²y-xy²)+xy(xy-x²)]/x²y=[2x^3y-2x^2y^2+x^2y^2-x^3y]/x²y=x-y=4
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-12=4x+y≥2√(4xy)=4√(xy)xy-4√(xy)-12≥0(√(xy)-6)(√(xy)+2)≥0√(xy)≤-2,√(xy)≥6因为√(xy)≥0所以√(xy)≥6xy≥36所以
(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^
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
(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60
(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22
7或者-8再问:求过程^_^再答:两个等式两边相加
2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3(x+y)+2xy=3*(-2)+2*3=0
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
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
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=
X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)
两式相加得到x+y=5,相减得y-x=1/5,故x=12/5,y=13/5xy=156/25,因为要求的都是正数,而且xy同正负,所以只考虑x,y正数即可故x²+y²=(x+y)^
X+y=3xy=-5X-y=?(x-y)^2=(X+y)^2-4xy=9+20=29则x-y=±根号29
x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式=9y²-3y²+y²/6y²=7/6x=4y时原式
[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+
x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60