方程组3又x 1=2y 2[x+1]-y=11
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x=log2(y)则X1+2X2+3X3=log2(y1)+2log2(y2)+3log2(y3)=log2(y1)+log2(y2^2)+log2(y3^3)=log2(y1y2^2y3^3)=1所
y^2-4x=0y^2-4y+2n=0有两不相同的解即:16-8n>0n再问:麻烦详细点,看不懂
由题α1+β1=5α2+β2=5α1β2+α2β1=α1(5-α2)+α2(5-α1)=5(α1+α2)-2α1α2又,方程组可化为关于x的一元二次方程3x²+5-x=29即,3x²
化简方程组得kx²-﹙2k+1﹚x+﹙k+½﹚=0{x=x1y=y1{x=x2y=y2为方程的实数解∵x=﹣b±﹙b²-4ac﹚/2a∴x1=2k+1/k+2y1=4k&
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+
y=(1-k^2)x-(k-2)经过点(0,-1)所以-1=(1-k^2)*0-(k-2),k=3所以直线方程是:y=-8x-1,随着x的增大而减小故y1>y2
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
两点都在直线上有(Y1-Y2)/(X1-X2)=2∴(x1-x2)+(y1-y2)=5(x1-x2)=16(x1-x2)=(x1+x2)-4*X1*X2方程组联立消Y得10X+12mX+3m+6=0由
①得,x=3-ky③代入②得,18-12ky+2k²y²+y²=6(2k²+1)y²-12ky+12=0∵两组实数解为(x1,y1)和(x2,y2),
根据题意:y1-y2=(1-2m)(1/x1-1/x2)=(1-2m)(x2-x1)/x1x20,x1x20,得到:m
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^2+y^2=m;x+y=2根据题意可得x1,x2是方程x^2+(2-x)^2=m即2x^2-4x+4-m=0的解所以x1+x2=-b/a=2x1*x2=c/a=(4-m)/2y1,y2是方程(2-
将y=3-x代入2x²+3y²=18得5x²-18x+9=0x=3或3/5所以y=0或12/5所以x1y2+x2y1=36/5第二题一式乘以三减去二式乘以二得y=(8-x
y=-2x+bk=-2,y随x的增大而减小∵x1
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+
x^2+y^2=4轨迹为半径为2的圆,在坐标系中画图,又OA垂直于OB,可知a=2或-2
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
y=-2x+5(x-y)²=1所以(3x-5)²=13x-5=±1x=4/3,x=2注意,不是x=-2所以x=4/3,y=7/3x=2,y=1
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|=√