\(\begin{aligned} & \sqrt[4]{32 \sqrt[3]{4}}+\sqrt[4]{64 \sqrt[3]{\frac{1}{2}}}-3 \sqrt[3]{2 \sqrt[4]{2}}=\sqrt[4]{2^5 \cdot 2^{2 / 3}}+\sqrt[4]{2^6 \cdot 2^{-1 / 3}}-3 \sqrt[3]{2 \cdot 2^{1 / 4}}= \\ & =2^{17 / 12}+2^{17 / 12}-3 \cdot 2^{5 / 12}=2 \cdot 2^{17 / 12}-3 \cdot 2^{5 / 12}=2^{5 / 12}(4-3)=2^{5 / 12}= \\ & =\sqrt[12]{2^5}=\sqrt[12]{32}\end{aligned}\)

\(\begin{aligned} & \frac{\sqrt[4]{7 \sqrt[3]{54}+15 \sqrt[3]{128}}}{\sqrt[3]{4 \sqrt[4]{32}}+\sqrt[3]{9 \sqrt[4]{162}}}=\frac{\sqrt[4]{7 \sqrt[3]{27 \cdot 2}+15 \sqrt[3]{64 \cdot 2}}}{\sqrt[3]{4 \sqrt[4]{16 \cdot 2}}+\sqrt[3]{9 \sqrt[4]{81 \cdot 2}}}=\frac{\sqrt[4]{7 \cdot 3 \sqrt[3]{2}+15 \cdot 4 \sqrt[3]{2}}}{\sqrt[3]{4 \cdot 2 \sqrt[4]{2}}+\sqrt[3]{9 \cdot 3 \sqrt[4]{2}}}= \\ & =\frac{\sqrt[4]{21 \sqrt[3]{2}+60 \sqrt[3]{2}}}{\sqrt[3]{8 \sqrt[4]{2}}+\sqrt[3]{27 \sqrt[4]{2}}}=\frac{\sqrt[4]{81 \sqrt[3]{2}}}{2 \sqrt[3]{\sqrt[4]{2}}+3 \sqrt[3]{\sqrt[4]{2}}}=\frac{3 \sqrt[4]{\sqrt[3]{2}}}{2 \sqrt[1]{2}+3 \sqrt[1]{2}}=\frac{3 \sqrt[12]{2}}{5 \sqrt[i 2]{2}}=\frac{3}{5} . \\ & \end{aligned}\)

\(\frac{\left(\frac{5}{8}+2 \frac{17}{24}\right): 2,5}{\left(1,3+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}} \cdot 0.5=\frac{\left(\frac{5}{8}+\frac{65}{24}\right) \cdot \frac{2}{5} \cdot \frac{1}{2}}{\left(\frac{13}{10}+\frac{23}{30}+\frac{4}{11}\right) \cdot \frac{110}{401}}=\frac{\frac{10}{3} \cdot \frac{1}{5}}{\frac{401}{165} \cdot \frac{110}{401}}=\frac{2}{3}: \frac{2}{3}=1\)

\(\Large \frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{\left(x+\sqrt{x^2-1}\right)^4-1}\).

\(\begin{aligned} & \text { ამოხსენი უტოლობა: }\left\{\begin{array}{l}\left(x+\sqrt{x^2-1}\right)^4-1 \neq 0, \\ x^2-1 \geq 0 .\end{array}\right. \\ & \frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{\left(x+\sqrt{x^2-1}\right)^4-1}=\frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{\left(x^2+2 x \sqrt{x^2-1}+x^2-1\right)^2-1}= \\ & =\frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{\left(2 x^2+2 x \sqrt{x^2-1}-1\right)^2-1}= \\ & =\frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{\left(2 x^2+2 x \sqrt{x^2-1}-1-1\right)\left(2 x^2+2 x \sqrt{x^2-1}-1+1\right)}= \\ & =\frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{\left(2 x^2+2 x \sqrt{x^2-1}-2\right)\left(2 x^2+2 x \sqrt{x^2-1}\right)}= \\ & =\frac{4 x\left(x+\sqrt{x^2-1}\right)^2}{2\left(x^2+x \sqrt{x^2-1}-1\right) 2 x\left(x+\sqrt{x^2-1}\right)}=\frac{x+\sqrt{x^2-1}}{\left(x^2-1\right)+x \sqrt{x^2-1}}= \\ & =\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}\left(\sqrt{x^2-1}+x\right)}=\frac{1}{\sqrt{x^2-1}} \text {. } \\ & \end{aligned}\)

\(\begin{aligned} & \frac{a^3-a-2 b-b^2 / a}{\left(1-\sqrt{\frac{1}{a}+\frac{b}{a^2}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^3+a^2+a b+a^2 b}{a^2-b^2}+\frac{b}{a-b}\right)= \\ & =\frac{\frac{a^4-a^2-2 a b-b^2}{a}}{\left(1-\sqrt{\frac{a+b}{a^2}}\right) \cdot(a+\sqrt{a+b})}:\left(\frac{a^2(a+1)+a b(a+1)}{(a-b)(a+b)}+\frac{b}{a-b}\right)= \\ & =\frac{a^4-\left(a^2+2 a b+b^2\right)}{(a-\sqrt{a+b})(a+\sqrt{a+b})}:\left(\frac{a(a+1)(a+b)}{(a-b)(a+b)}+\frac{b}{a-b}\right)= \\ & =\frac{a^4-(a+b)^2}{a^2-a-b}:\left(\frac{a(a+1)}{a-b}+\frac{b}{a-b}\right)= \\ & =\frac{\left(a^2-a-b\right)\left(a^2+a+b\right)}{a^2-a-b}: \frac{a^2+a+b}{a-b}= \\ & =\frac{\left(a^2+a+b\right)(a-b)}{a^2+a+b}=a-b=23-22=1\end{aligned}\)

უტოლობა: \(\left\{\begin{array}{l}a>0, \\ b>0 .\end{array}\right.\) \[ \begin{aligned} & \frac{\left(\sqrt[5]{a^{4 / 3}}\right)^{3 / 2}}{\left(\sqrt[5]{a^4}\right)^3} \cdot \frac{\left.\sqrt{a^{a^2 b}}\right)^4}{(\sqrt[4]{a \sqrt{b}})^6}=\frac{\left(a^{4 / 15}\right)^{3 / 2}}{\left(a^{4 / 5}\right)^3} \cdot \frac{\left.\sqrt{a \cdot a^{2 / 3} \cdot b^{1 / 3}}\right)^4}{\left(\sqrt[4]{a \cdot b^{1 / 2}}\right)^6}=\frac{a^{(4 / 15)(3 / 2)}}{a^{(4 / 5) 3}} \times \\ & \times \frac{\left(\sqrt{a^{1+2 / 3} \cdot b^{1 / 3}}\right)^4}{\left(a^{1 / 4} \cdot b^{1 / 8}\right)^6}=\frac{a^{2 / 5}}{a^{12 / 5}} \cdot \frac{\left(a^{5 / 6} \cdot b^{1 / 6}\right)^4}{a^{6 / 4} \cdot b^{6 / 8}}=a^{2 / 5-12 / 5} \cdot \frac{a^{(5 / 6) 4} \cdot b^{(1 / 6) 4}}{a^{3 / 2} \cdot b^{3 / 4}}= \\ & =a^{-10 / 5} \cdot \frac{a^{10 / 3} \cdot b^{2 / 3}}{a^{3 / 2} \cdot b^{3 / 4}}=a^{-2} \cdot a^{10 / 3-3 / 2} \cdot b^{2 / 3-3 / 4}=a^{-2} \cdot a^{(20-9) / 6} \cdot b^{(8-9) 12}= \\ & =a^{-2} \cdot a^{11 / 6} \cdot b^{-1 / 12}=a^{-2+1 / 6} \cdot \frac{1}{b^{1 / 12}}=\frac{a^{(-12+11) 6}}{b^{1 / 12}}=\frac{a^{-1 / 6}}{b^{1 / 12}}= \\ & =\frac{1}{a^{1 / 6} \cdot b^{1 / 12}}=\frac{1}{\sqrt[12]{a^2 b}} . \end{aligned} \]