方程组x 2=5.3x-2y=7

两式相减得：2x(x2-x1)+2y(y2-y1)+x1^2-x2^2+y1^2-y2^2=A-B得：y=kx+t,这里k=(x1-x2)/(y2-y1),t=(A-B-x1^2+x2^2-y1^2+

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

化简（1）得y(x-1)=2(x-1)然后分类讨论（因为要看x-1能不能直接两边同除）当x=1时,根据（2）式算出y=0；当x不等于1时y(x-1)=2(x-1)就可以两边同除以(x-1)得y=2,代

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

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

x=66/37y=-28/37

3x²+2y=6···········①y=m+x·············②②代入①,得：3x²+2(m+x)=63x²+2x+2m-6=0要使原方程组仅有一组实数解,

x^2+y^2-2x=0kx-y-k=0,y=kx-k代入x^2+(kx-k)^2-2x=0x^2+k^2x^2-2k^2x+k^2-2x=0(k^2+1)x^2+(-2k^2-2)x+k^2=0△=

y=x+my²=x²+2mx+m²=2xx²+(2m-2)x+m²=0韦达定理x1+x2=-(2m-2)x1x2=m²1/x1+1/x2=(

(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

题目正确吗?

∵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

x2+y2+2xy=9(x+y)²=9x+y=3或-3(x-y)2-3x+3y+2=0(x-y)²-3(x-y)+2=0(x-y-1)(x-y-2)=0x-y=1或2所以组成4个方

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

联立这两个方程,消去y得10x²+12mx+3m²+6=0(记为方程3)x1,x2为方程3的两个根,所以,方程3的判别式必须＞0,得144m²-4×10×（3m²

y1=x1+2,y2=x2+2,(y1-y2)=(x1-x2)√[(x1—x2)^2+(y1—y2)^2]=√2|x1-x2|连立代入有:x^2-2x+3=x+2x^2-3x+1=0|x1-x2|=√

x2+y2-4x-3y+5=0①3x2+3y2-11x-7y+10=0②3*①,得3x2+3y2-12x-9y+15=0③②-③,得x+2y=5x=5-2y④将④代入①,整理得y2-3y+2=0解之,

展开x^2-2y^2-xy=-5x^2-2y^2+xy=7所以x^2-2y^2=1xy=6上式除以下式,解分式方程x/y=3/2或-4/3(舍去)所以x^2=9y^2=4x=3y=2或x=-3y=-2

证明：∵kx-y-k=0∴y=kx-k代入x²+y²-2x=0得,（k²+1)x²-(2k²+2)x+k²=0∴x1+x2=(2k²