变量X~P(2)x1,x2-xn是来自X的样本,则DX-搜答案-智慧作业帮

拆开,得x^2-2x-mx+2m=p^2-2p-mp+2m移项得x^2-p^2-2x+2p-mx+mp=0(x-p)(x+p)-2(x-p)-m(x-p)=0(x-p)(x+p-2-m)=0x1=p,

一元二次方程ax^2+bx+c=0根与系数关系x1+x2=-b/ax1x2=c/a这就是韦达定理前提是方程的根存在

x1*x2=a-1x1+x2=-（a-2）因为点P(x1,x2)在圆x²+y²=4上所以x1²+x2²=4即（x1+x2）²-2x1*x2=4所以（a

x1+x2=-px1*x2=2(x1-x2)^2=(x1+x2)^2-4*x1*x2=p^2-8=4p=2根号（3）或-2根号（3）

α+β=-2（m+2）αβ=m²+3α²+β²=（α+β）²-2αβ=2m²+16m+10（α-1）²+(β-1)²=α²

原式可写为x^2-mx-2x+2m-(p-2)(p-m)=0在写为x^2-(2+m)x-p(p+2+m)=9用十字相乘法可得(x+p)(x-p-2-m)=0解得x1=-p,x2=p+2+m

x1x2x3x3x1x2x2x3x1c1+c2+c3x1+x2+x3x2x3x1+x2+x3x1x2x1+x2+x3x3x1r2-r1,r3-r1x1+x2+x3x2x30x1-x2x2-x30x3-

x1.x2是方程2x²-x-3=0的两实根∴x1+x2=1/2x1x2=-3/2∴x1+x2+x1*x2=1/2-3/2=-1

由题意得p(x,√（4ax))所以轨迹为y=√(4ax)即y=2√ax

用韦达定理因为X2大于X1所以x2-x1=p+1>0因为x1+x2=-(b/a)=-2x1*x2=c/a=p^2又因为（x1+x2)^2-4x1x2=(x2-x1)^2(这步可能不好理解,就慢慢拆开吧

分析：设P（x1,y1）,欲求出动点P的轨迹方程,只须求出x,y的关系式即可,结合新定义运算,即可求得动点P(x^2,4ax)的轨迹方程,从而得出其轨迹．∵x1*x2=(x1+x2)^2-(x1-x2

韦达定理呀,x1+x2=-Q,x1x2=-Px1^2+x2^2=(x1+x2)^2-2x1x2=Q^2+2P=71/x1+1/x2=(x1+x2)/x1x2=3∴Q=3P然后带进去算-（根号下77）/

x²-x+p-1=0的两个树根为x1,x2则x1+x2=1,x1x2=p-1(x1²-x1-2)(x2²-x2-2)=(x1x2)²-x1²x

1）方程x^2+mx+n=0(n≠0)的两根为x1.x2,且x1+x2=-m,x1*x2=n新方程的两根为y1,y2,y1+y2=1/x1+1/x2=(x1+x2)/x1*x2=-m/ny1*y2=1

观察得到：x1=p是方程的一个根.又x1+x2=2+m所以x2=m-p+2

1.（x-2)(x-m)-(p-2)(p-m)=0x^2-p^2-(2+m)x+(2+m)p=0(x-p)(x+p)-(2+m)(x-p)=0(x-p)[x+p-(2+m)]=0x1=px2=2+m-

1）x1^2/4+y1^2/2=1,①x2^2/4+y2^2/2=1,②①-②,(x1+x2)(x1-x2)/4+(y1+y2)(y1-y2)/2=0,x1+x2=2,∴(y1-y2)/(x1-x2)

3x²-7x+2=0)3x-1)(x-2)=0x1=1/3,x2=2所以x1+x2=7/3