x2次減3xy加p=
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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.
:(1)2x2-5x+x2+4x,其中x=-3=3x²-x=3x(-3)²+3=27+3=30(2)(3x2-xy-2y2)-2(x2+xy-2y2),其中x=6,y=-1=3x&
∵2(x2+xy)-3(xy+y2)=2x2-xy-3y2,∴2x2-xy-3y2=2×3-3×(-2)=12.故选A.
A、原式=-6x2-19xy-5y2;B、原式=2x2-9xy-7y2;C、原式=x2-16xy-10y2;D、原式=8x2-13xy-15y2.故选D.
∵x2+xy=-3,xy+y2=7,∴(x2+xy)+(xy+y2)=-3+7=4,即x2+2xy+y2=4.
x^2+y^2+2x-8y+17=0(x+1)^2+(y-4)^2=0x=-1y=4x的2005次+xy=-1-4=-5
5x2—3xy—4y2—11xy—7x2+2y2=-2x2-2y2-14xy=-2(x2+y2)-14xy=-14+28=14
x+y=7xy=27^2=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2*2x^2+y^2=45(1)y=7-xx(7-x)=2x^2-7x+2=0x=(7+41^(1/2))/2或x=(
再问:你确定吗?再答:我保证再答:亲,那一步不明白可以问再问:你省略了一步吧!再答:能看明白吗再问:no再答: 再答:是不是这部再问:那么第一步是怎么得来的呢?再答:第一步不是你
[(3x的2次幂y+2xy的2次幂)除以xy]的2次幂=(3x+2y)²=9x²+12xy+4y²
有x2+xy=3可得,2x2+2xy=6 (1),有xy+y2=-2得,3xy+3y2=-6 (2),根据分析,(1)-(2)可得,2x2-xy-3y2=6-(-6)=
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.
答:x²+y²=10xy=3,y=3/x代入上式得:x²+(3/x)²=10整理得:(x²)²-10*x²+9=0(x²
∵x2+y2=7,xy=-2,∴7x2-3xy-2y2-11xy-5x2+4y2,=(7x2-5x2)+(4y2-2y2)-(3xy+11xy),=2x2+2y2-14xy,=2(x2+y2)-14x
=[x+(3-√5)/2*y][x-(3-√5)/2*y]有点牵强,但这是唯一的答案了
2x²-xy-3y²=2x²+2xy-3xy-3y²=2(x²+xy)-3(xy+y²)已知x²+xy=3,xy+y²=
0.X^2+XY=3;(1)XY+Y^2=2;(2)2*(1)-3*(2)=2*X^2-XY-3*Y^2=2*3-3*2=0
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
原式=2x2-2xy-6x2+9xy-2x2+2x2-xy+y2=-2x2+5xy+2y2,当x=-1,y=-2时,原式=-2+10+8=16.