{x y=5003.7%x 2.25%y=15.6,怎么算
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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
∵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.
x2-y2=(x2-xy)+(xy-y2)=21-12=9;x2-2xy+y2=(x2-xy)-(xy-y2)=21+12=33.
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
∵x2+xy=2,y2+xy=5,∴x2+2xy+y2=7,则原式=12(x2+2xy+y2)=72,故答案为:72
X2+xy-(xy+y2)=4-12x2+xy-xy-y2=-8x2-y2=-8x2+xy+xy+y2=4+12x2+2xy+y2=16
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)=
x^2+xy+y^2=2≥3xyxy≤2/3-2xy≥-4/3,x^2-xy+y^2=x^2+xy+y^2-2xy=2-2xy≥2/3当且仅当x=y时取等号再问:>=2/3且
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.
∵x2+y2=7,xy=-2,∴7x2-3xy-2y2-11xy-5x2+4y2,=(7x2-5x2)+(4y2-2y2)-(3xy+11xy),=2x2+2y2-14xy,=2(x2+y2)-14x
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
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.
(1)∵x2-y2=x2+xy-xy-y2=x2+xy-(xy+y2)而x2+xy=2,xy+y2=-1,∴x2-y2=2-(-1)=3;(2)∵x2+2xy+y2=x2+xy+xy+y2,而x2+x