已知(x 2)与(y 2)2互为相反数求x的2010次方 y的2011次方的值

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已知(x 2)与(y 2)2互为相反数求x的2010次方 y的2011次方的值
已知圆C1:x2+y2+2x+2y-8=0与圆C2:x2+y2-2x+10y-24=0

(1)C1:(X+1)^2+(Y+1)^2=10圆心o1(-1,-1)C2:(X-1)^2+(Y+5)^2=50圆心o2(1,-5)O1O2^2=2^2+4^2=20

已知圆O:x2+y2=1与直线l:y=kx+2

(1)当k=2时,直线l的方程为:2x-y+2=0-------(1分)设直线l与圆O的两个交点分别为A、B过圆心O(0,0)作OD⊥AB于点D,则OD=|2×0-0+2|22+(-1)2=25---

已知│x-y+1│与x2+8x+16互为相反数,求x2+2xy+y2的值.

已知│x-y+1│与x2+8x+16互为相反数,求x2+2xy+y2的值.x-y+1=0;(1)(x+4)=0;x=-4;y=-4+1=-3;∴x2+2xy+y2=(x+y)²=(-4-3)

已知|x-y+1|与x2+8x+16互为相反数,求x2+2xy+y2的值.

∵|x-y+1|与x2+8x+16互为相反数,∴|x-y+1|与(x+4)2互为相反数,即|x-y+1|+(x+4)2=0,∴x-y+1=0,x+4=0,解得x=-4,y=-3.当x=-4,y=-3时

已知x,y∈R,比较x2+y2与2(2x-y)-5的大小

因为:x2+y2-{2(2x-y)-5}=x方+y方-4x+2y+5=(x方-4x+4)+(y方+2y+1)=(x-2)方+(y+1)方>=0如果x=2,y=-1时前者=后者其它的均是前者>后者

已知x2-xy=21,xy-y2=-12,分别求式子x2-y2与x2-2xy+y2的值.

x2-y2=(x2-xy)+(xy-y2)=21-12=9;x2-2xy+y2=(x2-xy)-(xy-y2)=21+12=33.

已知2x=3y,求xy/(x2+y2)-y2/(x2-y2)的值

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*

已知(x2+y2+3)(x2+y2-2)-6=0,求x2+y2的值

(x²+y²)²+(x²+y²)-6-6=0(x²+y²)²+(x²+y²)-12=0(x²

已知圆x2+y2-4ax+2ay+20(a-1)=0若该圆与圆x2+y2=4相切,求A的值

现将方程化简(x-2a)2+(y+a)2=-20(a-1)+5a2由相切得两圆心距离之和等于半径之和即(2a)2+a2={2+根号下5(a-2)2}2解得A=根号下(1\5)+1或-根号下(1\5)+

已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少

X2+xy-(xy+y2)=4-12x2+xy-xy-y2=-8x2-y2=-8x2+xy+xy+y2=4+12x2+2xy+y2=16

已知y=y1+y2,y1与根号x成正比例,y2与x2成反比例

由题意,不妨设:y1=k*根号x,y2=m/x²那么:y=y1+y2=k*根号x+m/x²已知当x=1时,y=-12;当x=4时,y=7,则有:{k+m=-12(1){2k+m/1

已知椭圆3x2+2y2=6x与曲线x2+y2-k=0恒有交点,求k的取值范围

x^2+y^2-k=0,y^=k-x^,①代入3x^+2y^=6x得3x^+2(k-x^)=6x,x^-6x+9=9-2k,当9-2k>=0②时x=3土√(9-2k),代入①得k-[3土√(9-2k)

已知x,y为实数,且(x2 +y2)(x2 +y2+2)=3.求x2 +y2的值

设t=x2+y2(t大于等于0)则t(t+2)-3=0(t+3)(t-1)=0t=-3(舍去)或t=1所以,x2+y2=1

已知x与y互为相反数,且x-y=5.求x2+y2的值

x^2+y^2=(x-y)^2+2xy而xy=[(x+y)^2-(x-y)^2]/4=-25/4所以原式=25-25/2=25/2或者根据x+y=0,直接求出x,y这样更简单

已知双曲线x2/a2-y2/b2=1的离心率为2,焦点与椭圆x2/25-y2/9=1的焦点相同,那么

双曲线离心力e=c/a=2椭圆x^2/25-y^2/9=1那是肯定不对的中间要不是加号,要不就双曲线!我就当+号来计算了即椭圆x^2/25+y^2/9=1椭圆半焦距c1=√(25-9)=4所以双曲线半

已知圆C1:x2+y2=10与圆C2:x2+y2+2x+2y−14=0.

(1)证明:圆C2:x2+y2+2x+2y−14=0化为标准方程为(x+1)2+(y+1)2=16∴C2(-1,1),r=4∵圆C1:x2+y2=10的圆心坐标为(0,0),半径为R=10∴|C1C2

若|m+4|与n2-2n+1互为相反数,把多项式x2+4y2-mxy-n分解因式.______.

由题意可得|m+4|+(n-1)2=0,∴m+4=0n−1=0,解得m=−4n=1,∴x2+4y2-mxy-n,=x2+4y2+4xy-1,=(x+2y)2-1,=(x+2y+1)(x+2y-1).

已知实数x.y满足(x2+y2)(x2+y2-1)=2,求x2+y2的值

可设x²+y²=t.则t(t-1)=2.===>t²-t-2=0.===>(t-2)(t+1)=0.===>t=2.即x²+y²=2.

1已知│x-y+1│与x2+8x+16互为相反数,求x2+2xy+y2的值

(1)|x-y+1|≥0,x^2+8x+16=(x+4)^2≥0两者互为相反数,所以|x-y+1|=x^2+8x+16=0x-y+1=0,x+4=0,所以x=-4,y=-3x2+2xy+y2=(x+y