Question

(1+t)/(t^2-2t+1),t>1且t≠0,的值域怎么求

Answer 1

(t+1)/(t²-2t+1)=(t+1)/(t-1)²

t>1 t+1>0 t-1>0

令f(t)=(t²-2t+1)/(t+1)

f(t)=(t²+t-3t-3+4)/(t+1)

=[t(t+1)-3(t+1)+4]/(t+1)

=t-3+4/(t+1)

=t+1+4/(t+1)-4

由均值不等式得t+1+4/(t+1)>4

f(x)>0

函数的值域为(0,+∞)