Question

设x1,x2是方程3x²-2x-4=0的两根,不解方程,求下列各式的值 ①1/x1+1/x2 ②x2/x1+x1/x2

③（x1-x2）² ④x1³+x2³

Answer 1

解：因为 x1, x2是方程 3x^2--2x--4=0的两根，
所以 x1+x2=2/3, x1*x2=--4/3,
所以
(1) 1/x1+1/x2=(x1+x2)/(x1*x2)
=(2/3)/(--4/3)
=--1/2.
(2) x2/x1+x1/x2=(x2^2+x1^2)/(x1*x2)
=[(x1+x2)^2--2x1*x2]/(x1*x2)
=[(2/3)^2--2(--4/3)]/(--4/3)
=[(4/9)+(8/3)]/(--4/3)
=(32/9)/(--4/3)
=--8/3.
(3) (x1--x2)^2=(x1+x2)^2--4x1*x2
=(2/3)^2--4(--4/3)
=4/9+16/3
=52/9.
(4) x1^3+x2^3=(x1+x2)^3--3x1*x2(x1+x2)
=(2/3)^3--3(--4/3)(2/3)
=8/27+8/3
=80/27。

Answer 2

x1+x2=1/3
x1*x2=-4/3

1.
=(1/3)/(-4/3)=-1/4

2.
=[(1/3)²/(-4/3)]-2=-25/12

3.
=(1/3)²-4*(-4/3)=49/9

4.
=(1/3)*[(1/3)²-3*(-4/3)]=37/27

Answer 3

路过......