\(I = 2\log _a^{\frac{1}{2}}b\left( {{{\left( {{{\log }_a}\sqrt[4]{{ab}} + {{\log }_b}\sqrt[4]{{ab}}} \right)}^{\frac{1}{2}}}-{{\left( {{{\log }_a}\sqrt[4]{{\dfrac{b}{a}}} + {{\log }_b}\sqrt[4]{{\dfrac{a}{b}}}} \right)}^{\frac{1}{2}}}} \right)\)
\({\left( {{{\log }_a}\sqrt[4]{{ab}} + {{\log }_b}\sqrt[4]{{ab}}} \right)^{\frac{1}{2}}} = {\left( {\frac{1}{4}\left( {{{\log }_a}a + {{\log }_a}b} \right) + \dfrac{1}{4}\left( {{{\log }_b}a + {{\log }_b}b} \right)} \right)^{\frac{1}{2}}} = \)
\( = {\left( {\dfrac{1}{4}\left( {1 + {{\log }_a}b + {{\log }_b}a + 1} \right)} \right)^{\frac{1}{2}}} = {\left( {\dfrac{1}{4}\left( {2 + {{\log }_a}b + \dfrac{1}{{{{\log }_b}b}}} \right)} \right)^{\frac{1}{2}}} = \)
\( = \sqrt {\dfrac{{\log _a^2b + 2{{\log }_a}b + 1}}{{4{{\log }_a}b}}} = \sqrt {\dfrac{{{{\left( {{{\log }_a}b + 1} \right)}^2}}}{{4{{\log }_a}b}}} = \dfrac{{\left| {{{\log }_a}b + 1} \right|}}{{2\sqrt {{{\log }_a}b} }}.\)
Так как по условию \(a > 1\) и \(b > 1,\) то \({\log _a}b > 0\) и \(\left| {{{\log }_a}b + 1} \right| = {\log _a}b + 1.\)
\({\left( {{{\log }_a}\sqrt[4]{{\dfrac{b}{a}}} + {{\log }_b}\sqrt[4]{{\dfrac{a}{b}}}} \right)^{\frac{1}{2}}} = {\left( {\dfrac{1}{4}\left( {{{\log }_a}b-1 + {{\log }_b}a-1} \right)} \right)^{\frac{1}{2}}} = \)
\( = {\left( {\dfrac{1}{4}\left( {{{\log }_a}b + \dfrac{1}{{{{\log }_a}b}}-2} \right)} \right)^{\frac{1}{2}}} = \sqrt {\dfrac{{\log _a^2b-2{{\log }_a}b + 1}}{{4{{\log }_a}b}}} = \sqrt {\dfrac{{{{\left( {{{\log }_a}b-1} \right)}^2}}}{{4{{\log }_a}b}}} = \dfrac{{\left| {{{\log }_a}b-1} \right|}}{{2\sqrt {{{\log }_a}b} }}.\)
Если \(1 < a \le b\), то \(\left| {{{\log }_a}b-1} \right| = {\log _a}b-1\) и
\(I = 2\sqrt {{{\log }_a}b} \cdot \left( {\dfrac{{{{\log }_a}b + 1}}{{2\sqrt {{{\log }_a}b} }}-\dfrac{{{{\log }_a}b-1}}{{2\sqrt {{{\log }_a}b} }}} \right) = 2\sqrt {{{\log }_a}b} \cdot \dfrac{2}{{2\sqrt {{{\log }_a}b} }} = 2.\)
Если \(1 < b < a\), то \(\left| {{{\log }_a}b-1} \right| = 1-{\log _a}b\) и
\(I = 2\sqrt {{{\log }_a}b} \cdot \left( {\dfrac{{{{\log }_a}b + 1}}{{2\sqrt {{{\log }_a}b} }}-\dfrac{{1-{{\log }_a}b}}{{2\sqrt {{{\log }_a}b} }}} \right) = 2\sqrt {{{\log }_a}b} \cdot \dfrac{{2{{\log }_a}b}}{{2\sqrt {{{\log }_a}b} }} = 2{\log _a}b.\)
Ответ: 2, если \(1 < a \le b;\) \(2{\log _a}b\), если \(1 < b < a.\)