Question

设方程4A*A-7A=3的两根为A1,A2不解方程求(A1-3)(A2-3)值,A1*A1*A2+A1*A2*A2值A2

Answer 1

A1+A2=7/4

A1A2=-3/4

(A1-3)(A2-3)=A1A2-3(A1+A2)+9=-3/4-21/4+9=3

A1*A1*A2+A1*A2*A2=A1A2(A1+A2)=-21/16

A2/(A1+1)+A1/(A2+1)=(A2^2+A2+A1^2+A1)/(A1A2+A1+A2+1)

=((A1+A2)^2-2A1A2+A1+A2)/(A1A2+A1+A2+1)=101/32

(A1-A2)^2=(A1+A2)^2-4A1A2=97/16

所以A1-A2=+-根号97/4