lim(x,y)-ox^2y x^3 y^3
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 03:28:17
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=
∵3x-5y=0,∴x=5y3,∴原式=5y3−2y5y3+3y=-111.
令y=ux则x^2(xdu+udx)/dx=2(ux)^2+ux^2约掉x^2(xdu+udx)/dx=2(u)^2+u所以(xdu)/dx=2(u)^2之后你该知道了吧求出u关于x的表达式再有y=u
∵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
根据题意,x>0,y>0,则2yx>0,8xy>0,则2yx+8xy≥22yx•8xy=8,即2yx+8xy的最小值为8,若2yx+8xy>m2+2m恒成立,必有m2+2m<8恒成立,m2+2m<8⇔
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
-x^2y+5xy^2-yx^2=-2x^2y+5xy^2
证明:用反证法.假设1+xy与1+y2都大于或等于2,即1+xy≥21+yx≥2,∵x,y∈R+∴1+x≥2y1+y≥2x两式相加,得x+y≤2,与已知x+y>2矛盾.所以假设不成立,即原命题成立.
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
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
xy+yx=10x+y+10y+x=11x+11y=100+x10x=100-11yx=10-1.1y所以y只能是0
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
解答如下:x+2y=(yx)/44x+8y=xyxy-8y=4x(x-8)y=4x当x≠8时(x=8不成立)y=4x/(x-8)x+2y=(2x+1)/32y=(2x+1)/3-x2y=(1-x)/3
根号下则x-2>=0,x>=22-x>=0,x
化简:[(y²-x²)/(5x²-4yx)]/[(x+y)/(5x-4y)]原式=[(y+x)(y-x)/x(5x-4y)]×[(x+y)/(5x-4y)]=(y-x)/