Question

Срочно!!!! Нужно очень...

Answer 1

Ответ:

$\displaystyle \frac{1}{a^{\frac{1}{4}}+a^{\frac{1}{8}}+1}+\frac{1}{a^{\frac{1}{4}}-a^{\frac{1}{8}}+1}-\frac{2\, a^{\frac{1}{4}}-2}{a^{\frac{1}{2}}-a^{\frac{1}{4}}+1}=\\\\\\=\frac{a^{\frac{1}{4}}-a^{\frac{1}{8}}+1+a^{\frac{1}{4}}+a^{\frac{1}{8}}+1}{(\, (a^{\frac{1}{4}}+1)+a^{\frac{1}{8}}\, )(\, (a^{\frac{1}{4}}+1)-a^{\frac{1}{8}})\, }-\frac{2\, a^{\frac{1}{4}}-2}{a^{\frac{1}{2}}-a^{\frac{1}{4}}+1}=$

$\displaystyle =\frac{2\, a^{\frac{1}{4}}+2}{(a^{\frac{1}{4}}+1)^2-(a^{\frac{1}{8}})^2}-\frac{2\, (a^{\frac{1}{4}}-1)}{a^{\frac{1}{2}}-a^{\frac{1}{4}}+1}=\\\\\\=\frac{2\, (a^{\frac{1}{4}}+1)}{a^{\frac{1}{2}}+2a^{\frac{1}{4}}+1-a^{\frac{1}{4}}}-\frac{2\, (a^{\frac{1}{4}}-1)}{a^{\frac{1}{2}}-a^{\frac{1}{4}}+1}=\frac{2\, (a^{\frac{1}{4}}+1)}{a^{\frac{1}{2}}+a^{\frac{1}{4}}+1}-\frac{2\, (a^{\frac{1}{4}}-1)}{a^{\frac{1}{2}}-a^{\frac{1}{4}}+1}=$

$\displaystyle =\frac{2\, (a^{\frac{1}{4}}+1)(a^{\frac{1}{2}}-a^{\frac{1}{4}}+1)-2(a^{\frac{1}{4}}-1)(a^{\frac{1}{2}}+a^{\frac{1}{4}}+1)}{(a^{\frac{1}{2}}+a^{\frac{1}{4}}+1)(a^{\frac{1}{2}}-a^{\frac{1}{4}}+1)}=\\\\\\=\frac{2\, (\, (a^{\frac{1}{4}})^3+1^3)-2\, (\, (a^{\frac{1}{4}})^3-1^3)}{(a^{\frac{1}{2}}+1)^2-(a^{\frac{1}{4}})^2}=\frac{2\, (\, a^{\frac{3}{4}}+1)-2\, (\, a^{\frac{3}{4}}-1)}{a+2a^{\frac{1}{2}}+1-a^{\frac{1}{2}}}=\frac{4}{a+a^{\frac{1}{2}}+1}$