←Vrati se na zadatke3. ZadatakPojednostavni izraz −x−x−x(x−xx+xx−xx−x−x+xxx)-x^{-x^{-x}}\left(\frac{x^{-x^x}+x^{x^{-x}}}{x^{-x^{-x}}+x^{x^x}}\right)−x−x−x(x−x−x+xxxx−xx+xx−x) za x>0x>0x>0.Prikaži rješenjeRješenjeSupstitucijom xx=ux^x=uxx=u dobivamo: −x−x−x(x−xx+xx−xx−x−x+xxx)=−x−1u⋅1xu+x1u1x1u+xu=−x−1u⋅1+xux1uxu1+xux1ux1u=−x−1u⋅x1uxu=1xu=−x−u=−x−xx\begin{aligned}-x^{-x^{-x}}\left(\frac{x^{-x^x}+x^{x^{-x}}}{x^{-x^{-x}}+x^{x^x}}\right)&=-x^{-\frac1u}\cdot\frac{\dfrac1{x^u}+x^{\frac1u}}{\dfrac1{x^{\frac1u}}+x^u}\\&=-x^{-\frac1u}\cdot\frac{\dfrac{1+x^u x^{\frac1u}}{x^u}}{\dfrac{1+x^u x^{\frac1u}}{x^{\frac1u}}}\\&=-x^{-\frac1u}\cdot\frac{x^{\frac1u}}{x^u}=\frac1{x^u}=-x^{-u}=-x^{-x^x}\end{aligned}−x−x−x(x−x−x+xxxx−xx+xx−x)=−x−u1⋅xu11+xuxu1+xu1=−x−u1⋅xu11+xuxu1xu1+xuxu1=−x−u1⋅xuxu1=xu1=−x−u=−x−xxPomakni formulu lijevo ili desno.