x y=1,xy=-12.求x的平方 y的平方和x-y的值
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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
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
=-x-(2y-2+3x)+2(x+4)=-x-2y+2-3x+2x+8=-4x-2y+10
即x=y=1xy=1对不对?如果对的话x^2+2y^2-2xy-2y+1=0化简为你做的很对.就是这样解的,没有其他更好的方法了.这里用到的是数学里的
-xy(x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10
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
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15
5xy+4x+7y+6x-3xy-4xy+3y=5xy-3xy-4xy+4x+6x+7y+3y=(5-3-4)xy+(4+6)x+(7+3)y=-2xy+10x+10y=-2xy+10(x+y)因为x
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
原式=3x-3y-6xy=3-12=-9
(x+1)^2+|y-1|=0(x+1)^2>=0|y-1|>=0x+1=0,y-1=0x=-1,y=12(xy-5xy的平方)-(3xy的平方-xy)=2xy-10xy^2-3xy^2+xy=3xy
=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
条件是(x+1)^2+|y-1|=0,是吗?如果是它的话,那么x=-1,y=1.带入原式,得值为2
全微分啊dz=(1+xy)^x[ln(1+xy)+xy/(1+xy)]dx+(1+xy)^xx^2/(1+xy)dy
(-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*
x²+xy=99x(x+y)=9*11x(x+y)=9*(9+2)x=9,y=2
3(x^2-xy)-xy+y^2=3(x^2-xy)-(xy-y^2)=3*5-(-3)=15+3=18
5x²+5xy+y²+2x+1=04x²+y²+4xy+2x+x²+1=0(2x+y)²+(x+1)²=0得{2x+y=0---①
(-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