从答案解析看,是(1+2t)⁴,不是(1+t)⁴。
(1+t)⁶=(1+2t)⁴
(1+t)⁶-(1+2t)⁴=0
[(1+t)³]²-[(1+2t)²]²=0
[(1+t)³+(1+2t)²带圆侍][(1+t)³-(1+2t)²]=0
(t³+3t²+3t+1+4t²+4t+1)(t³腔肢+3t²+3t+1-4t²-4t-1)=0
(t³+7t²+7t+2)(t³-t²-t)=0
t(t³+7t²+7t+2)(t²-t-1)=0
t≠0,t³+7t²+7t+2单蠢吵调递增,令t=-½,得t³+7t²+7t+2=1/8>0
t³+7t²+7t+2恒>0
因此只有t²-t-1=0
t²=t+1