已知方程组12x y=10y 2等于

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已知方程组12x y=10y 2等于
解方程组 ﹛x2+y2=25和xy=-12

还有一种情况没照出来x=-4y=3

已知实数x,y满足方程组{x+xy+y=2+3倍根号2 {x2+y2=6 求(x+y+1)的绝对值

2x+2xy+2y=4+6根2x2+y2+2x+2xy+2y=10+6根号2(x+y)2+2(x+y)=10+6根号2(x+y+1)2=9+6根号2=(2根号2+1)平方所以x+y+1的绝对值是2根号

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*

解关于x,y的方程组{x2-y2+根号(x2+y2)=a xy=0

由xy=0,得x=0,或y=0当x=0时,代入方程1:-y^2+根号y^2=a,即y^2-|y|+a=0,解得|y|=[1±√(1-4a)]/2当y=0时,代入方程1:x^2+根号x^2=a,即x^2

已知x2+xy=2,y2+xy=5,则12x2+xy+12y2=___.

∵x2+xy=2,y2+xy=5,∴x2+2xy+y2=7,则原式=12(x2+2xy+y2)=72,故答案为:72

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x2+xy=5,xy+y2=-1,则x2-y2=______.

∵x2+xy=5,xy+y2=-1,∴(x2+xy)-(xy+y2)=x2+xy-xy-y2=x2-y2=5-(-1)=6.故填:6

已知X2-2x+y2+6y+10=0,求(x2-2xy)/(xy+y2)的值

x²-2x+y²+6y+10=0,变换得(x-1)²+(y+3)²=0,∴x=1,y=-3∴(x2-2xy)/(xy+y2)=(1²-2*(-3))/

已知xy满足x2+y2-6x+2y+10=0,求立方根号x2-y2的值

条件变换:(x-3)^2+(y+1)^2=0即:y+1=0x-3=0所以:立方根号x2-y2=2

已知实数x,y满足方程组x+xy+y=2+3根号2和x2+y2=6.求|x+y+1|的值.

(X+Y+1)^2=X^2+Y^2+1+2XY+2X+2Y=(X^2+Y^2)+2(X+XY+Y)+1=6+2*(2+3√2)+1=6+4+6√2+1=11+6√2

解方程组 x2+y2=10 x+xy+y=7

∵x^2+y^2=10,x+xy+y=7∴7-xy=x+y,且2xy≤x^2+y^2=10,∴(7-xy)^2=(x+y)^2=x^2+y^2+2xy=10+2xy,且xy≤5∴49-14xy+(xy

解方程组x2+y2=10,xy=3

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

解方程组 x2+y2=4 xy-y2+4=0

第一个式子:x的平方等于4-y的平方第二个式子:4-y的平方等于-xy说明x=-y所以x=根号2;y=-根号2或则x=-根号2;y=根号2

解方程组x2+y2=20,2x2-3xy-2y2=o

由2x²-3xy-2y²=0得2-3(y/x)-2(y/x)²=0(2+y/x)*(1-2y/x)=0得y/x=1/2或-2即y=1/2x或y=-2x代入x²+

解方程组x2+xy=12 xy+y2=4

x^2+xy=12xy+y^2=4因式分解下,得x(x+y)=12.y(x+y)=4两个方程相加,得(x+y)^2=16所以x+y=±4当x+y=4时,代入x(x+y)=12.y(x+y)=4解得x=

解二元二次方程组①x2-y2=3②x2+y2+2xy+x+y=12方程组如上:求详解

因为X^2-Y^2=(X+Y)(X-Y)x^2+y^2+2xy=(X+Y)^2这个题目可以因式分解成1,(X+Y)(X-Y)=32,(X+Y)^2+(X+Y)=12设X+Y=AX-Y=B那么方程变成A

已知实数x,y满足x2+xy+y2=3,则x2-xy+y2的最小值

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

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²