若实数xy满足x2十y2 小于等于1
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/06 02:48:19
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
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
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~
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
由基本不等式得x2+y2>=2根号(x2y2)=2丨xy丨,即2丨xy丨
设x=rsina,y=rcosa1=-1故3r^2/2>=u>=r^2/26>=u>=1/2
∵xy+yz+zx≤x2+y22+y2+z22+x2+z22=x2+y2+z2=1,又∵2(xy+yz+zx)=(x+y+z)2-(x2+y2+z2)≥0-1=-1,∴xy+yz+zx≥−12.故选B
∵x2+y2=1,∴x=sinθ,y=cosθ,∴(1-xy)(1+xy)=1-x2y2=1-(sinθcosθ)2=1-(12sin2θ)2=1-14sin22θ,当sin2θ=0时,1-14sin
∵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.
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=
再问:好的再答:求采纳
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
观察到sin²θ+cos²θ=1,则可做三角代换令x=sinθ,y=cosθ(1-xy)(1+xy)=1-(xy)²=1-(sinθcosθ)²=1-(sin2
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)/
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
∵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
由x2+y2=1得到的是|xy|=1-1/4=3/4没有用到两边平方吧,你说的两边平方指什么?
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.
再问:лл再问:ʮ�ָ�л
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