x~N(0,4^2)x1,x2--x20一组计算P

因为x1、x2是方程2X^2-2x+3m-1=0的根所以x1+x2=-(-2/2)=1x1*x2=(3m-1)/2又x1*x2/(x1+x2-4)

选DA错：X1=0,时,X2=0成立,X2=2时,挂掉!B错：P=1C错：D=0,1,2,3

由韦达定理x1+x2=-3x1x2=n3x1-x2=4所以x2=3x1-4代入x1+x2=-34x1-4=-3x1=1/4x2=3x1-4=-13/4所以n=x1x2=-13/16

X1＋X2=-B/A=2X1*X2=C/A=1/2求得X1=1+根号2或者X1=1-根号2从而求出X2的值X1/X2+X2/X1=(X1*X1+X2*X2)/(X1X2)=6

唯达定理：x1+x2=2,x1x2=1/2→x1²+x2²=(x1+x2)²-2x1x2=3→x1/x2+x2/x1=(x1²+x2²)/x1x2=6

∵X1、X2是方程X²-4X+2=0的两个根∴X1+X2=4,X1·X2=2X2/X1+X1/X2=X2²/(X1·X2)+X1²/(X1·X2)=(X1²+X

2x²+4x+1=0的两个根为x1,x2带入2(x1)²+4(x1)+1=02(x2)²+4(x2)+1=0粮食相减得2[(x1)²-(x2)²]+4

根据韦达定理有X1+X2=-b/a=-2/3,X1*X2=c/a=-3/3=-1①x2/x1+x1/x2=(x2²+x1²)/（x1x2）=【（x1+x2）²-2x1x2

方程4x^2-7x-3=0的两根为x1,x2,所以x1+x2=7/4,x1x2=-3/4,x2/(x1+1)+x1/(x2+1)=（x1^2+x2^2+x1+x2）/(x1x2+x1+x2+1)x1^

1方程x^2+4x+3=0的两个根为x1=?,x2=?.x1+x2=?,x1*x2=?x²+4x+3=0(x+1)(x+3)=0x=-1或x=-3x1=-1,x2=-3,x1+x2=-4,x

这是韦达定理x1+x2=-3/4x1x2=-2x1+x2=把根求出来才能得出记得采纳啊

x²-2x-4=0由根与系数关系知：x1+x2=2,x1x2=-4x2/x1+x1/x2=(x2²+x1²)/x1x2=[(x1+x2)²-2x1x2]/x1x

二次方程的根与系数关系..由已知得...x1+x2=-mx1*x2=n由x1与x2的范围可以知道x1+x2属于[0,正无穷大)...而x1*x1属于(负无穷,正无穷)...所以m属于(负无穷,0]n为

N(0,σ^2)E(X1+X2)=EX1+EX2=0D(X1+X2)=DX1+DX2=2σ^2X1+X2~N(0,2σ^2)同理：X1-X2~N(0,2σ^2)所以1/√2σ(X1+X2)~N(0,1

题目写清楚点儿啊X1+X2=-3/2X1*X2=-2|X1-X2|=√41/2析：由根与系数的关系即得X1+X2=-3/2与X1*X2=-2而|X1-X2|^2=（X1+X2）^2-4X1*X2m=-

对于一元二次方程ax2+bx+c=0,若存在根x1、x2,则x1+x2=-b/a,x1*x2=c/a；对于本题,x1+x2=4/3,x1*x2=-2/3,所以(1)=（x1+x2）^2-2x1*x2=

由x1≤0及0≤x2≤1∴x1+x2=m²+n²-6n≤1（1）x1×x2=m²+n²+2m-4n+1≤0（2)由（2）（m²+2m+1）+n&sup

韦达定理x1+x2=4x1x2=2所以1/x1+1/x2=(x1+x2)/x1x2=2

x1^2=3-x1,x2^2=3-x2,x1^3-4x2^2+19=3x1-x1^2-4x2^2+19=3x1-x1^2+4x2+7,x1+x2=-1,原式=4+x2-x1^2=4+x2-3+x1=0

因为正态分布具有再生性,就是由这些样本经过变形组成的样本空间,仍然服从正态分布N(2,4),则E(X)=2,D(X)=4则E[(X1+X2+X3+X4)/4]=1/4[E(X1)+E(X2)+E(X3