Question

谁能帮我解一道二次根式的题?(根号3,+1)的2001次方-2(根号3,+1)的2000次方-2(根号3.+1)的199

Answer 1

(根号3,+1)的2001次方-2(根号3,+1)的2000次方-2(根号3.+1)的1999次方+2001

设S=根号3+1

则原式=s^2001-2s^2000-2s^1999+2001

=s^1999(s^2-2s-2)+2001

=s^1999[(s-1)^2-3]+2001

=s^1999*[(根3+1-1）^2-3]+2001

=s^1999[3-3]+2001

=s^1999*0+2001

=2001