题目

解个方程:(3+t2-2t-3)2+t2=(t-1)2+(4+t2-2t-3)2
t2都是t的平方,以此类推!

提问时间：2020-12-22

答案

(3+t2-2t-3)2+t2=(t-1)2+(4+t2-2t-3)2
(t2-2t)2+ t2=(t-1)2+( t2-2t+1)2
(t2-2t)2+ t2=(t-1)2+[( t2-2t)+1]2
(t2-2t)2+ t2=(t-1)2+( t2-2t)2+2( t2-2t)+1
t2=(t-1)2+2( t2-2t)+1
t2= t2-2t+1+2t2-4t+1
2t2-6t+2=0
t2-3t+1=0
利用求根公式即可求出方程的根