x2(x-y) y2(y-x)怎么算
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因为方程可以变成(x+2)2+(y-3)2=0所以x+2=0且y-3=0,所以xy=-6
∵x2-4x+y2+6y+z−3+13=0,∴(x-2)2+(y+3)2+z−3=0,∴x-2=0,y+3=0,z-3=0,解得x=2,y=-3,z=3,∴(xy)z=[2×(-3)]3=-216.
首先,X^2-XY+2Y^2=(X-2Y)(X+Y)所以,设x^2-xy-2y^2-x+5y-2可分解成(X-2Y+A)(X+Y+B)则展开有X的一次项=A+B=-1Y的一次项有A-2B=5连列成方程
当x,y≥0时,曲线x2+y2=|x|+|y|互为x2+y2=x+y,曲线表示以(12,12)为圆心,以22为半径的圆,在第一象限的部分;当x≥0,y≤0时,曲线x2+y2=|x|+|y|互为x2+y
x,y∈R,且x^2+y^2
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*
x^2+xy-2y^2-x+7y-6=(x+2y)(x-y)-x+7y-6x+2y-3Xx-y2十字相乘得:2(x+2y)-3(x-y)=-x+7y是一次项所以原式=(x+2y-3)(x-y+2)
将x2+y2+174=4x+y,变形得:(x2-4x+4)+(y2-y+14)=0,即(x-2)2+(y-12)2=0,解得:x=2,y=12,则原式=2×122+12=25.
设k=x+y,y=k-x代入得:x^2+(k-x)^2=12x^2-2kx+k^2-1=0方程有解,判别式>=0,则有:4k^2-8(k^2-1)>=0-4k^2+8>=0k^2
[(x^2+y^2)-(x-y)^2+2y(x-y)]÷4y=1(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)÷4y=1(4xy-2y^2)4y=12x-y=24x/(4x^2-y^2)
x2-2xy+y2+3x-3y+2=(x-y)2+3(x-y)+2=(x-y-1)(x-y-2).
x²+y²-10x-6y+34=0,即(x-5)²+(y-3)²=0,所以,x=5,y=3,所以(x+y)/y=(5+3)/3=8/3
对x求导0.5*1/(x²+y²)*(x²+y²)'=1/[1+(y/x)²]*(y/x)'0.5/(x²+y²)*(2x+2y*
(X+Y)²=X²+Y²+2XY=X²+Y²+X²-Y²=2X²(X-Y)²=X²+Y²-
第一个圆的半径为1,第二个圆的半径为4,而圆心C1(0,0)C2(3,4)的距离为d=√(3-0)^2+(4-0)^2=5,所以两圆外切,有三条共切线,其中一条为内共切线过两圆切点,两圆直接相减即得方
解题思路:先根据去括号法则去括号,再合并同类项,最后代入数值进行计算。解题过程:
不对=x2(x-y)-y2(x-y)=(x2-y2)(x-y)=(x+y)(x-y)2再问:噢。我看懂了
∵(x-2)2+(y+3)2+z+2=0,∴x-2=0,y+3=0,z+2=0,解得x=2,y=-3,z=-2,∴(xy)z=(-6)-2=136.
哥!你那个是x方y方吧!有这么个公式x方-y方=(x+y)(x-y)所以得到了(x+y)(x-y)-(x+y)这时候提取公因式(x+y)就得到了(x+y)(x-y-1)再问:是啊,怎么提(X+Y)他那
(x-y)/(x+y)=(x-y)(x+y)/[(x+y)^2]=(x^2-y^2)/[x^2+y^2+2xy]=2xy/[x^2+y^2+x^2-y^2]=2xy/(2x^2)=y/xx^2-y^2