已知,·y x=3 4
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∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,(x-1)2+(y-2)2=0,∴x=1,y=2,∴yx−xy=2-12=1.5;故答案为:1.5.
方程x2+y2-4x+1=0表示以点(2,0)为圆心,以3为半径的圆.设yx=k,即y=kx,由圆心(2,0)到y=kx的距离为半径时直线与圆相切,斜率取得最大、最小值,由|2k−0|k2+1=3,解
yx+xy=xyx+xyy=xy(x+y)xy,∵x+y=8,xy=6,∴原式=6×86=463.
∵x2+y2-6x-8y+25=0,∴(x-3)2+(y-4)2=0,∴x=3,y=4,当x=3,y=4时,原式=43-34=712.
∵3x-5y=0,∴x=5y3,∴原式=5y3−2y5y3+3y=-111.
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)
∵1-8x≥0,8x-1≥0,∴x=18,y=12,∴代数式xy+yx+2-xy+yx−2=14+4+2-14+4−2=52-32=1.故选:B.
x−yx÷(x−2xy−y2x)=x−yx÷(x2x−2xy−y2x)=x−yx÷x2−2xy+y2x=x−yx÷(x−y)2x=x−yx•x(x−y)2=1x−y,当x=2009,y=2010时,原
原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y,由已知得(3x-2y)(x+y)=0,因为x+y≠0,所以3x
要使根号3-x和根号x-3有意义则3-x=0x=3算出y=2yx次方:2³=8立方根为2最后结果为2
似乎题目应该是y=√(x-2)+√(2-x)+4x-2>=02-x>=0x=2代入得y=4yx=4*2=8y的x次=4²=16
(1)x=4代入,4+y=4y-2y;所以y=4(2)y=4代入,4+4=4x-2*4;所以x=4
∵x2-4xy+4y2=0,∴(x-2y)2=0,∴x=2y,∴x-yx+y=2y-y2y+y=13.故分式x-yx+y的值等于13.
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
两边对x求导得y'x^2+2xy+2y^3+4xy^2y'=0解出来y'就可以了再问:4xy^2y'为什么是4xy再答:搞错了,应该是6xy^2y'再问:yx^2+2xy^3+3=-18上点(1,-2
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.
yx+xy=x2+y2xy=(x+y)2-2xyxy当x+y=2,xy=-5时,原式=22-2×(-5)-5=-145.故答案为-145.
2x2-xy-3y2=0,(2x-3y)(x+y)=0,解得:2x-3y=0或x+y=0(分母为0,舍去),解得:x=3y2,则x−yx+y=3y2−y3y2+y=y5y=15.
∵5x2-xy-6y2=0,∴(5x-6y)(x+y)=0,∴5x-6y=0,x+y=0,∴5x=6y,x=-y,∴yx=56或-1.故答案为:56或-1.
即(10x+y)*(10y+x)=2268101xy+10x²+10y²=2268因为后面的10x²+10y²只可能是整十的数,所以2268中的个位8要靠101