已知x y=3,xy=1,试求 (1)x² y²得值
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/05 08:30:09
xy^2*(-x^2y^2)*(1/2xy^3)^2=xy^2*(-x^2y^2)*(1/4x²y^6)=xy^2*(x^4y^8)*(-1/4)=(-1/4)*xy²*(xy
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
绝对值大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0所以x+y+1=0xy+3=0xy=-3x+y=-1两边平方x^2+2xy+y^2=(-1)^2x^2+y^2=1-
-xy(x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10
(-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
解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(
平方和绝对值都大于等于0相加等于0,若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以x+1=0,y-1=0x=-1,y=1后面漏了平方吧原式=2xy-2xy²-3xy+xy&su
∵x2+xy=-3,xy+y2=7,∴(x2+xy)+(xy+y2)=-3+7=4,即x2+2xy+y2=4.
答案是3/2你是不是把分母打错了教你个方法因为上下是齐次的直接令x=3y=2带入就行
代数式可化为2x+2xy-xy+3y+x+xy=3x+3y+2xy=3(x+y)+2xy=3×(-2)+6=0a^2-b^2=15 a^2-2ab+b^2=5&nbs
3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9
x2+2xy-3y2=x2-xy+3xy-3y2=x2-xy+3(xy-y2),∵x2-xy=3,xy-y2=-5,∴x2+2xy-3y2=3+3×(-5)=-12.
3xy=2x+3y+5(3y-2)(x-1)=7所以3y-2=7x-1=1得y=3x=2xy=6或3y-2=1x-1=7得y=1x=6xy=6所以xy=6
3(x^2-xy)-xy+y^2=3(x^2-xy)-(xy-y^2)=3*5-(-3)=15+3=18
原式=-xy²(x²y^4-xy²-1)∵xy²=-2原式=2((-2)²-(-2)-1)=10
(-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