若实数xy满足(x2=y2)2 x2-y2-2=0

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已知x、y是实数且满足x2+xy+y2-2=0,设M=x2-xy+y2,则M的取值范围是______.

由x2+xy+y2-2=0得:x2+2xy+y2-2-xy=0,即(x+y)2=2+xy≥0,所以xy≥-2;由x2+xy+y2-2=0得:x2-2xy+y2-2+3xy=0,即(x-y)2=2-3x

实数x、y满足x2+xy+y2=2,记u=x2-xy+y2,则u的取值范围是

x2+xy+y2=(x+y)2-xy=2,所以(x+y)2=2+xy.2|xy|+xy≤x2+xy+y2=2,所以0≤xy≤2/3.或者-2≤xy≤0u=x2-xy+y2=(x+y)2-3xy=2-2

已知实数XY 满足X2+Y2=1,求Y+2/X+1的取值范围

x^2+y^2=1代表圆C令(x+2)/(y+1)=k题目转化为圆上一点到点A(-2,-1)的斜率的取值范围连接AC,作垂直于AC交圆两点E,F画出图易得到k的取值咯说啥也要满足你E(-1,0),F(

若实数x、y满足方程x2+y2+3xy=35,则xy的最大值为

x^2+y^2>=2xy5xy再问:?再答:怎么了不对么再问:x^2+y^2>=2xy为什么?再答:(x-y)^2≥0x^2-2xy+y^2≥0x^2+y^2≥2xy这下理解了吧~

若实数XY满足X2+Y2=1,则X-2Y的最大值为

设:S=x-2y,x=S+2y代入x²+y²=1中得:(s+2y)²+y²=15y²+4sy+s²-1=0∵y是实数,∴△≥0(4s)&su

已知实数x,y满足(x2)+(y2)-xy+2x-y+1=0,求x,y的值

x²+(2-y)x+y²-y+1=0方程有解的条件是:△=b²-4ac≥0→-3y²≥0∴y=0∴x=-1

设实数x,y满足x2+y2=1则 xy的取值范围是?

由基本不等式得x2+y2>=2根号(x2y2)=2丨xy丨,即2丨xy丨

实数xy满足x2+y2+2x-4y+1=0,则x2+y2-2x+1的最小值为

x^2+y^2+2x-4y+1=0(x+1)^2+(y-2)^2=2^2x=-1+2cosx,y=2+2sinxy-x=3+2(sinx-cosx)=3+2√2(sinx/√2-cosx/√2)=3+

若实数x,y满足3x2+2y2=6x,则x2+y2的最大值为______.

∵3x2+2y2=6x,∴y2=-32x2+3x,由y2=-32x2+3x≥0,可得0≤x≤2,又x2+y2=x2-32x2+3x=-12x2+3x=-12(x-3)2+92,∵0≤x≤2,∴x=2时

若实数x,y满足|xy|=1,则x2+4y2的最小值为______.

∵x2+4y2≥2x2•4y2=4|xy|=4,当且仅当|x|=2|y|=2时取等号,∴x2+4y2的最小值为4.故答案为:4.

设正实数x,y满足x2-xy+y2=1,求x2-y2的最大值和最小值

令:x=a+b,y=a-bx^2-xy+y^2-1=0==>a^2+3*b^2=1,a=sinT,b=(√3)(cosT)/3x^2-y^2=4ab=(2√3)(sin2T)/3>0因此:最小值0=

已知实数x、y满足2x2-7xy+3y2=0,求x:y

分解因式有(x-3y)(2x-y)=0所以有x=3y或2x=y所以x:y=3:1或x:y=1:2

已知实数x、y满足X2+y2-xy+2x-y+1=0,试求x、y的值

x^2+(2-y)x+y^2-y+1=0这个关于x的二次方程有解b^2-4ac>0-3y^2>0所以y=0x=-1

已知非零实数x,y满足:x2+xy-2y2=0,求(x2+3y+y2)/(x2+y2)的值

由x2+xy-2y2=0,x2-y2+xy-y2=0,(x+y)(x-y)+(x-y)y=0,(x-y)(x+2y)=0.得x-y=0或者x+2y=0.1)当x-y=0,x=y.(x2+3y+y2)/

设实数xy满足x2 +2y2=6,则x+y的取值范围

解由x2+2y2=6得x2/6+y2/3=1故设x=√6cosa,y=√3sina则x+y=√6cosa+√3sina=3(√6/3cosa+√3/3sina)=3sin(a+θ)由-3≤3sin(a

若实数x,y满足x2+y2+xy=1,则x+y的最大值是 (  )

∵实数x,y满足x2+y2+xy=1,即(x+y)2=1+xy.再由xy≤(x+y)24,可得(x+y)2=1+xy≤1+(x+y)24,解得(x+y)2≤43,∴-43≤x+y≤43,故 

已知实数x.y满足(x2+y2)(x2+y2-1)=2,求x2+y2的值

可设x²+y²=t.则t(t-1)=2.===>t²-t-2=0.===>(t-2)(t+1)=0.===>t=2.即x²+y²=2.

若实数x,y满足x2-2xy+y2+x-y-6=0,则x-y的值是(  )

设x-y=m,则原方程可化为:m2+m-6=0,解得m1=2,m2=-3,所以,x-y的值2或-3.故选B.

已知实数xy满足x2+y2=3,若m=y+1/x+3

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已知实数x,y满足x2+xy+y2=3,则x2-xy+y2的最小值

由x2+xy+y2=3得,x^2+y^2=3-xyx^2+y^2≥2xy得,xy≤1所以x^2-xy+y^2=3-2xy≥1等号成立当且仅当x=y=±1