如果f(t)=t/(1+t),g(t)=t/(1-t) ,证明:f(t)-g(t)=-2g(t²)
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如果f(t)=t/(1+t),g(t)=t/(1-t) ,证明:f(t)-g(t)=-2g(t²)
如果f(t)=t/(1+t),g(t)=t/(1-t) ,证明:f(t)-g(t)=-2g(t²)
如果f(t)=t/(1+t),g(t)=t/(1-t) ,证明:f(t)-g(t)=-2g(t²)
f(t)-g(t)
=t/(1+t)-t/(1-t)
=t/(1+t)+t/(t-1)
=[t(t-1)/(1+t)(t-1)]+[t(t+1)/(t+1)(t-1)]
=(t²-t+t²+t)/(t+1)(t-1)
=2t²/(t²-1)
-2g(t²)
=-2t²/(1-t²)
=2t²/(t²-1)
左边等于右边
所以f(t)-g(t)=-2g(t²)成立
得证
f(t)-g(t)=t/(1+t)-t/(1-t)
=-2t²/[(1+t)(1-t)]
=-2[t²/(1-t²]=-2g(t²)
f(t)=t/(1+t)
g(t)=t/(1-t)
f(t)-g(t)
=t/(1+t)-t/(1-t)
=[t(1-t)-t(1+t)]/(1+t)(1-t)
=(t-t^2-t-t^2)/(1-t^2)
=-2t^2/(1-t^2)
-2g(t^2)
=-2t^2/(1-t^2)
所以
f(t)-g(t)= -2g(t2)
证明: g(t^2)=t^2/(1-t^2)
f(t)-g(t)=t/(1+t)-t/(1-t)
=[t*(1-t) - t*(1+t)]/(1-t)(1+t)
=-2t^2/(1-t^2)
=-2g(t^2)
证毕
同上
代入即可
左边=t/(1+t)-t/(1-t)=t(1/(t+1)+1/(t-1))=2t^2/(t^2-1)=-2t^2/(1-t^2)=右边