若丨x-2丨 (3y 2)²
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原式=[2(2x+y)]2-3x2-y2+2-2y2=4(2x+y)2-3(x2+y2)+2∵2x+y=7,x2+y2=5∴原式=183.
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*
y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得:y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以
y1=y2-2/3x+1=x2/3x+x=15/3x=1x=3/5y1=y2-5-2/3x+1=x-52/3x+x=1+55/3x=6x=18/5
∵3x2+2y2=6x,∴y2=-32x2+3x,由y2=-32x2+3x≥0,可得0≤x≤2,又x2+y2=x2-32x2+3x=-12x2+3x=-12(x-3)2+92,∵0≤x≤2,∴x=2时
(4x-3y)2-16y2=[(4x-3y)+4y][(4x-3y-4y]=(4x+y)(4x-7y).
x2-2xy+y2+3x-3y+2=(x-y)2+3(x-y)+2=(x-y-1)(x-y-2).
由y1-y2=x^2-2x-3=(x-3)(x+1),得1)当x>3orxy22)当-1
即(x²+4x+4)+(y²-6y+9)=0(x+2)²+(y-3)²=0所以x+2=y-3=0x=-2,y=3所以原式=(4+4)/(4+27)=8/31
手机不行,希望你能看明白吧,主要考察我们对于圆锥曲线跟圆的知识
可以看到没有根号时,那两个分别是以(—3,0)和(3,0)为圆心的圆,即条件要求两个圆的相交点正好半径和等于10.根据两圆关于y轴对称时正好可以得到一个特殊点(0,4)或者(0,—4)满足条件.所以最
解由当x=2时,y1+y2=-1得2k1+2k2=-1即k1+k2=-1/2.①当x=3时,y1-y2=12得3k1-3k2=12即k1-k2=4.②由(1)与(2)联立解得k1=7/4,k2=-9/
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
由题意可得,y2=3x−3x22由y2≥0可得3x−3x22≥0解可得,0≤x≤2设t=x2+y2=x2+3x−3x22=−12x2+3x=−12(x2−6x)=−12(x−3)2+92∵0≤x≤2又
1.y1=y2:-x+2=3x+4=>4x=-2=>x=-1/2y1-1/2y1>y2:-x+2>3x+4=>xx=1y=1带入y=ax+7=>a=-6
x-2=3(2x-3)x-2=6x-96x-x=-2+95x=7x=1.4
3x2+2y2-6x=0x2+y2=1/2(6x-x2)=9/2-1/2(x2-6x+9)=9/2-2-1/2(x-3)2当x=3时,Z最大=4.5
2x²+xy-3y²+x+4y-1=2×x×x+x×y-3×y×y+x+4×y-1=x×2x+x×y-y×3y+x×1+y×4-1=x×(2x+y+1)-y×(3y+4)-1再问:
原式=x2+6xy+9y2+2x4-2y4+x2-6xy+9y2=2x2+18y2+2x4-2y4
【-(-2x^2y)^3】^2*(-3x^3y2)=(8x^6*y³)²*(-3x³y²)=64x^12*y^6*(-3x³y²)=-192