已知x1和x2是方程x^2+6x+7的两根,求下列各式的值.1）（x1-x2)^2 2)x1/x2+x2/x1

已知x1和x2是方程x^2+6x+7的两根,求下列各式的值.1）（x1-x2)^2 2)x1/x2+x2/x1
已知x1和x2是方程x^2+6x+7的两根,求下列各式的值.1）（x1-x2)^2 2)x1/x2+x2/x1

已知x1和x2是方程x^2+6x+7的两根,求下列各式的值.1）（x1-x2)^2 2)x1/x2+x2/x1
由韦达定理
x1+x2=-6,x1*x2=7
1）（x1-x2)^2
=x1^2-2x1x2+x2^2
=x1^2+2x1x2+x2^2-4x1x2
=(x1+x2)^2-4x1x2
=(-6)^2-4*7
=8
2)x1/x2+x2/x1
=(x1^2+x2^2)/x1x2
=(x1^2+2x1x2+x2^2-2x1x2)/x1x2
=[(x1+x2)^2-2x1x2]/x1x2
=[(-6)^2-2*7]/7
=22/7

x^2+6x+7=0
x1+x2=-6 x1x2=7
(x1-x2)^2=(x1+x2)^2-4x1x2=36-28=8
x1/x2+x2/x1=(x1^2+x2^2)/x1x2=[(x1+x2)^2-2x1x2)]/x1x2
=(36-14)/7=22/7