求x^2y x^3 y^3极限
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/12 15:35:07
根号下大于等于0所以x-2>=0,x>=22-x>,x
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=
∵x+y=4,xy=3,∴原式=x2+y2xy=(x+y)2−2xyxy=16−63=103.
∵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)
4xy加5x²y加yx²再问:谢谢了,
x=6-3y &nbs
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
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
合并同类项3-2xy+3yx^2+nxy-4x^2y=3+(n-2)xy+3yx^2-4x^2y不含XY项n-2=0n=2
3x^2y-3xy^2+6yx^2-9y^2x=3x^2y+6yx^2-3xy^2-9y^2x=9x^2y-12xy^2=3xy(3x-4y)
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
因为y=3xy+x,所以x-y=-3xy,当x-y=-3xy时,2x+3xy−2yx−2xy−y=2(x−y)+3xy(x−y)−2xy=2(−3xy)+3xy−3xy−2xy=35.
∵|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.
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.
您好:-3y^2-2yx+x^2=(x-3y)(x+y)如果本题有什么不明白可以追问,如果满意请点击右下角“采纳为满意回答”如果有其他问题请采纳本题后另发点击向我求助,答题不易,请谅解,谢谢.祝学习进
根号下则x-2>=0,x>=22-x>=0,x
6x²y+2xy-3x²y²-7x-5yx-4y²x²-6x²y|x+3|+(y+1/2)²=0x=-3y=1/2=-3xy-7x