设 A=12x2−3xy−y2 , B=2x2−4xy−2y2
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![设 A=12x2−3xy−y2 , B=2x2−4xy−2y2](/uploads/image/f/7243484-68-4.jpg?t=%E8%AE%BE+A%3D12x2%E2%88%923xy%E2%88%92y2+%2C+B%3D2x2%E2%88%924xy%E2%88%922y2)
由题意可得a=x2−xy+y2x2+xy+y2=(xy)2−xy+1(xy)2+xy+1,令xy=t≠0,可得a=t2−t+1t2+t+1=1-2tt2+t+1=1-2t+1t+1,变形可得21−a−
∵A=x2+xy+y2,B=-3xy-x2∴2A-3B=2(x2+xy+y2)-3(-3xy-x2)=2x2+2xy+2y2+9xy+3x2=5x2+11xy+2y2.
设x2-xy+y2=Ax2-xy+y2=A与x2+xy+y2=1相加可以得到:2(x2+y2)=1+A(1)x2-xy+y2=A与x2+xy+y2=1相减得到:2xy=1-A(2)(1)+(2)×2得
∵A=2x2-3xy+y2+x-3y,B=4x2-6xy+2y2+4x-y,∴a=B-2A=(4x2-6xy+2y2+4x-y)-2(2x2-3xy+y2+x-3y)=4x2-6xy+2y2+4x-y
设x^2-xy+y^2=Ax^2-xy+y^2=A与x2+xy+y2=3相加可以得到2(x^2+y^2)=3+A(1)x^2-xy+y^2=A与x2+xy+y2=3相减得到2xy=3-A(2)(1)+
A=B+C=(2x2-3xy-y2)+(x2+xy+y2)=2x2-3xy-y2+x2+xy+y2=3x2-2xy.故选D.
(1)-X2(2)-3ab-9bc+6ac(你打错了吧,应该是2ab-3bc-ac吧)(3)-4
∵x<y<0,∴x-y<0,x+y<0.∴x2−2xy+y2=(x−y)2=|x-y|=y-x.x2+2xy+y2=(x+y)2=|x+y|=-x-y.∴x2−2xy+y2+x2+2xy+y2=-2x
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
由xy=0,得x=0,或y=0当x=0时,代入方程1:-y^2+根号y^2=a,即y^2-|y|+a=0,解得|y|=[1±√(1-4a)]/2当y=0时,代入方程1:x^2+根号x^2=a,即x^2
(1)2A-B=2(3x2+3y2-5xy)-(2xy-3y2+4x2)=6x2+6y2-10xy-2xy+3y2-4x2=2x2+9y2-12xy;(2)当x=3,y=−13时,2A-B=2x2+9
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.
设x2-xy+y2=M①,x2+xy+y2=3②,由①、②可得:xy=3−M2,x+y=±9−M2,所以x、y是方程t2±9−M2t+3−M2=0的两个实数根,因此△≥0,且9−M2≥0,即(±9−M
x^2+xy=12xy+y^2=4因式分解下,得x(x+y)=12.y(x+y)=4两个方程相加,得(x+y)^2=16所以x+y=±4当x+y=4时,代入x(x+y)=12.y(x+y)=4解得x=
4xy-(2x2+5xy-y2)+2(x2+3xy),=4xy-2x2-5xy+y2+2x2+6xy),=5xy+y2,∵.x+2 .+(y-12)2=0∴x=-2,y=12当x
(3A-2B)-(2A+B)=3A-2B-2A-B=A-3B,将A、B代入,即得:4x2-4xy+y2-3(x2+xy-5y2)=4x2-4xy+y2-3x2-3xy+15y2=x2-7xy+16y2
=[x+(3-√5)/2*y][x-(3-√5)/2*y]有点牵强,但这是唯一的答案了
2x2+(-x2-2xy+2y2)-3(x2-xy+2y2)=-2x2+xy-4y2 当x=2,y=-12时,原式=-10.
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