方程组﹛4.5x 2.5y=36 15x 25y=36﹜
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mx+y=3y=3-mx代入x^2+2y=5x^2+2(3-mx)-5=0x^2-2mx+1=0方程组有一组实数解,所以△=0(-2m)^2-4*1*1=0m^2=1m=±1m=1时,x=1m=-1时
5x+y=369y-x=2即45y-10+y=36y=1x=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+
令x/4=y/5=K,则x=4k,y=5k.所以12k-10k=10.解得K=5.则x=4k=20,y=5k=25.所以x=20,y=25再问:可不可以不要不看题就复制啊,你再做一次吧再答:因为x/4
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
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△=
y1=2x1+m,y2=2x2+m(x1-x2)+(y1-y2)=(x1-x2)+(2x1+m-2x2-m)=5(x1-x2)=6,∴(x1-x2)=6/5联立y=2x+m,x/3-y/2=110x+
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
由x-y=k得,x=y+k代入x^2+y^2=16中得(y+k)^2+y^2=16整理得,2y^2+2ky+k^2-16=0∵有解∴△≥0即4k^2-8(k^2-16)≥0解得-4√2≤k≤4√2
令x/4=y/5=K,则x=4k,y=5k.所以12k-10k=10.解得K=5.则x=4k=20,y=5k=25.所以x=20,y=25
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
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,
解用图像来做方程x²+y²=4表示以(0,0)为圆心,2为半径的圆方程x+y=a表示直线圆心(0,0)到直线x+y=a的距离为d=/a//√2当d=2,即/a//√2=2,即a=±