Question

设x1,x2,是方程2x的平方-9x+6=0的两个根

Answer 1

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