Question

设x1,x2,是方程2x的平方-9x+6=0的两个根

（1）x1分之一+x2分之一 （2）x1的平方+x2的平方 （3）（X1-3)(X2-3) （4）x1-x2

Answer 1

解：由韦达定理可知x1+x2=-2分之9,x1x2=3
1）原式=(x1x2)分之(x1+x2)
=3分之(-2分之9)
=-2分之3
2）原式=(x1+x2)²-2x1x2
=(-2分之9)²-2×3
=4分之81-6
=4分之57
3）原式=x1x2-3(x1+x2)+9
=3-3×(-2分之9)+9
=3+2分之27+9
=2分之51
4）①x1>x2
则原式=√[(x1+x2)²-4x1x2]
=√(4分之81-12)
=√(4分之33)
=2分之√33
②x1<x2
同理可求得x1-x2=-2分之√33
所以x1-x2=±2分之√33

Answer 2

x1+x2=9/2 x1x2=3
(1)1/x1+1/x2=(x1+x2)/(x1x2)=(9/2)/3=3/2
(2)1/x1²+1/x2²=(x1²+x2²)/(x1x2)²=[(x1+x2)²-2x1x2]/(x1x2)²=[81/4-6]/9=19/12
(3)(x1-3)(x2-3)=x1x2-3(x1+x2)+9=3-27/2+9=-3/2
(4)(x1-x2)²=(x1+x2)²-4x1x2=81/4-12=33/4，∴x1-x2=√33/2或-√33/2