已知x y=-4,xy=-12,求代数式x 1分之y 1 y 1分之x 1的值
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x2-xy=-3①,2xy-y2=-8②,①×2+②×3得:2x2-2xy+6xy-3y2=-6-24=-30,则2x2+4xy-3y2=-30.
∵x2-xy=7,2xy+y2=4,∴原式=(x2-xy)+(2xy+y2)=7+4=11,故答案为:11
原式=3xy+6y-xy+6x=2xy+6(x+y)=2*(-2)+6*4=24-4=20
(4xy+12y)+(7x-(3xy+4Y-x))=4xy+12y+(7x-3xy-4y+x)=4xy+12y+7x-3xy-4y+x=(4-3)xy+(12-4)y+(7+1)x=xy+8y+8x当
xy-12=4x+y≥2√(4xy)=4√(xy)xy-4√(xy)-12≥0(√(xy)-6)(√(xy)+2)≥0√(xy)≤-2,√(xy)≥6因为√(xy)≥0所以√(xy)≥6xy≥36所以
因为x-y=4xy所以x-2xy-y=2xy2x+3xy-2y=11xyx-2xy-y分之2x+3xy-2y=5.5
原式=[4(x+y)-2xy]分之[(x+y)+xy]=[4(3xy)-2xy]分之[(3xy)+xy]=10xy分之2xy=5分之1
1、已知2x²+xy=10,3y²+2xy=6,求4x²+8xy+9y²的值为?分析:通过观察,可以把8xy拆成2xy+6xy,分别于剩余的两项组合,并提取公因
(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8
把已知两式相加结果为13
3(xy+2y)-(xy-6x)=3xy+6y-xy+6x=2xy+6(x+y)=2×(-2)+6×4=-4+24=20
∵x2+xy=2,y2+xy=5,∴x2+2xy+y2=7,则原式=12(x2+2xy+y2)=72,故答案为:72
xx+2xy-yy=-3两式相加即可
X2+xy-(xy+y2)=4-12x2+xy-xy-y2=-8x2-y2=-8x2+xy+xy+y2=4+12x2+2xy+y2=16
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
xy+1/xy>=2√(xy*1/xy)=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√(x/y*y/x)=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求
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=
x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式=9y²-3y²+y²/6y²=7/6x=4y时原式