Rationalize $$C=\dfrac{\sqrt{t}}{\sqrt[4]{t}+\sqrt{t}}$$
We have $D_t:t>0.$ I did the following: $$C=\dfrac{\sqrt{t}}{\sqrt[4]{t}+\sqrt[4]{t^2}}\cdot\dfrac{\sqrt[4]{t}-\sqrt[4]{t^2}}{\sqrt[4]{t}-\sqrt[4]{t^2}}=\dfrac{\sqrt{t}\left(\sqrt[4]{t}-\sqrt{t}\right)}{\sqrt{t}-t}\cdot\dfrac{\sqrt{t}+t}{\sqrt{t}+t}=\\=\dfrac{\sqrt{t}(\sqrt[4]{t}-\sqrt{t})(\sqrt{t}+t)}{t-t^2},t\ne1.$$
When $t=1$ $$C=\dfrac{1}{1+1}=\dfrac{1}{2}$$
The given answer in my book is $$C=\dfrac{(\sqrt[4]{t}-\sqrt{t})(1+\sqrt{t})}{1-t}.$$ Where am I wrong, or how can I further simplify my answer? Is this a reasonable approach? Thank you!