(3x 1-x 1)÷x的平方-4x 4x 1,从-1≤x≤4中选一个整数代入

X1,X2是方程2x的平方+3x-4=0的两个实数根x1+x2=-3/2x1x2=-2x1^2+2x1x2+x^2=9/4x1^2-2x1x2+x^2=9/4-4x1x2(x1-x2)^2=41/4x

1.k=1或者-4分子32.-1好简单哦,两道题都用那个什么定理来回答!就是x1+x2=-b/a、x1*x2=c/a

(X1)³-4(X2)²+19=(X1)*(X1)²-4(X2)²+19=(X1)*(-X1+3)-4(-X2+3)+19=-(X1)²+3(X1)+

由韦达定理得：X1+X2=-2,X1*X2=-3/2,X1^2+X2^2=(X1+X2)^2-2X1*X2=4+3=7(X2/X1)^2+(X1/X2)^2=［(X1^2+X2^2)^2-2(X1*X

x1+x2=5x1x2=31/x1+1/x2=(x1+x2)/(x1x2)=5/3x1²+x2²=(x1+x2)²-2x1x2=19

根据韦达定理x1+x2=-3/2,x1x2=-2所以x1²+x2²=(x1+x2)²-2x1x2=(-3/2)²+4=9/4+4=25/4

设x1>x2,x1+x2=-1对方程变形：x1^2+x1-3=0x1^3+x1^2-3x1=0x2^2+x2-3=0所以x1^3-4x2^2+19=3x1-x1^2-4(3-x2)+19=3x1-(3

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

∵x²+6x+3=0∴x1+x2=-6x1x2=3x1/x2+x2/x1=(x1+x2)²-2x1x2/x1x2=10

x1,x2是方程的解,所以带入方程得x1²-4×x1+k-3=0(1)x2²-4×x2+k-3=0(2)∵x1=3x2∴代入（1）得9x2²-12×x2+k-3=0(3)

X的平方-3X+1=0的两个实数根是X1,X2X1+X2=3X1X2=1(X1-X2)^2=(X1+X2)^2-4X1X2=3^2-4=5X1-X2=正负根号5

题目不完整,x1^2+x2^2=什么,没有这个条件做不出,希望你能补全题目再问：已知X1,X2是方程2x的平方+3x-4=0的两个实数根,求x1的平方+x2，用韦达定理再答：x1²+x2&s

x-x+3=0所以x1+x2=1,x1x2=3因此(1)(X1+2)(X2+2)=x1x2+2(x1+x2)+4=3+2x1+4=9(2)(X1-X2)=(x1+x2)-4x1x2=1-4x3=-11

方程3x²-4x=-1可化为：3x²-4x+1=0由根与系数的关系,有x1+x2=4/3,x1x2=1/3∴x2/x1+x1/x2=(x1²+x2²)/(x1x

设x1,x2是方程2x平方+4x-3=0的两个根,则x1+x2=-2x1·x2=-3/2∴x1平方+x2平方=(x1+x2)²-2x1·x2=(-2)²-2×(-3/2)=4+3=

3x^2+4x-7=0由韦达到理得：x1+x2=-4/3、x1x2=-7/3.x1^2+x2^2=(x1+x2)^2-2x1x2=16/9+14/3=58/9.1/x1^2+1/x2^2=(x1^2+

1x1\3=1/2*(1/1-1/3)2x1\4=1/2*(1/2-1/4).1x1\3+2x1\4+3x1\5+.+2006x1\2008=1/2(1/1-1/3+1/2-1/4+1/3-1/5+.

解.由韦达定理知,x1+x2=-k;x1x2=4k²-3.代入到x1+x2=x1x2中.得-k=4k²-3.解得k=-1,k=3/4.

X1X2＝1X1+X2＝3x1^2-4x1-x2=x1^2-4x1-(3-x1)=x1^2-3x1-3∵x1,x2是方程x^2-3x+1=0的解∴x1^2-3x1+1-4=-4

x1+x2=4x1x2=-1(x1+x2)^2/(1/x1+1/x2)=(x1+x2)^2*x1x2/(x1+x2)=x1x2*(x1+x2)=-4