在计算x+y x-3y
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依相反数的意义有|x-y+3|=-|x+y-1999|.因为任何一个实数的绝对值是非负数,所以必有|x-y+3|=0且|x+y-1999|=0.即x−y+3=0①x+y−1999=0②,由①有x-y=
[23x−2x+y(x+y3x−x−y)]÷x−yx=[23x−2x+y•x+y3x+2x+y•(x+y)]•xx−y=(23x−23x+2)•xx−y=2xx−y故答案为2xx−y.
[2x(x²y-xy²)+xy(xy-x²)]/x²y=[2x^3y-2x^2y^2+x^2y^2-x^3y]/x²y=x-y=4
∵x+y=4,xy=3,∴原式=x2+y2xy=(x+y)2−2xyxy=16−63=103.
设点P(x,y)为圆(x-3)2+(y-3)2=6的动点,则yx=kOP为原点O与P点连线的斜率,运动点P,可得当直线OP与圆C相切时,OP的倾斜角达到最值,同时斜率也达到最值.设经过原点的圆的切线为
∵3x-5y=0,∴x=5y3,∴原式=5y3−2y5y3+3y=-111.
∵x−yx+y=2,∴x-y=2(x+y),∴x−yx+y-2x+2yx−y=2(x+y)x+y-2(x+y)2(x+y)=2-1=1,故答案为:1.
2x+yx2-2xy+y2•(x-y)=2x+y(x-y)2•(x-y)(2分)=2x+yx-y;(4分)当x-3y=0时,x=3y;(6分)原式=6y+y3y-y=7y2y=72.(8分)
4xy加5x²y加yx²再问:谢谢了,
x=6-3y &nbs
y=√(x-2)+√(2-x)(x-2)+√(2-x)+3√(x-2)>=0x>=2√(2-x)>=0x
满足约束条件的平面区域如下图所示:联立x=yx+2y=3可得x=1y=1.即A(1,1)由图可知:当过点A(1,1)时,2x-y取最大值1.故答案为:1
3x^2y-3xy^2+6yx^2-9y^2x=3x^2y+6yx^2-3xy^2-9y^2x=9x^2y-12xy^2=3xy(3x-4y)
原式=7x²y+2yx²-2xy²+y²x-3xy²=9x²y-4xy²3x²y-5xy²=-2x²
xy+yx=10x+y+10y+x=11x+11y=100+x10x=100-11yx=10-1.1y所以y只能是0
(1):-2X-3Y+5-7Y+3X-5=(-2+3)X+(-3-7)Y+(5-5)=X-10Y(2):5X^3+2X^2Y+8X^3-4YX^2+3Y-1=(5+8)X³+(2-4)X
∵|x-y+3|+|x+y-1999|=0,∴x-y+3=0,x+y-1999=0,即x-y=-3,x+y=1999,则x+yx−y=1999−3=-19993.
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
由题意得,x−y=2x−2y+3=3,解得:x=4y=2,则可得a=3,b=2,b-a=-1,-1的立方根为:-1.
画出满足约束条件(x-2)2+y2=3的几何图形如下图示:∵yx−4表示(4,0)点于点P连线斜率,又由于圆的半径为3,易得过(4,0)点作圆的切线斜率分别为:±3故yx−4的最小值等于-3故答案为: