Cho hàm số \(f\left( x \right) = {\left( {{x^{1 + \dfrac{1}{{2{{\log }_4}x}}}} + {8^{\dfrac{1}{{3{{\log }_{{x^2}}}2}}}} + 1} \right)^{\dfrac{1}{2}}} - 1\)  với \(0 < x \ne 1\). Tính giá trị biểu thức \(P = f\left( {f\left( {2018} \right)} \right)\).

Ta có:

\(\begin{array}{l}{x^{1 + \dfrac{1}{{2{{\log }_4}x}}}} = {x^{1 + \dfrac{1}{{{{\log }_2}x}}}} = {x^{1 + {{\log }_x}2}} = {x^{{{\log }_x}2x}} = 2x\\{8^{\dfrac{1}{{3{{\log }_{{x^2}}}2}}}} = {2^{3.\dfrac{1}{{3{{\log }_{{x^2}}}2}}}} = {2^{\dfrac{1}{{{{\log }_{{x^2}}}2}}}} = {2^{{{\log }_2}{x^2}}} = {x^2}\end{array}\)

Khi đó \(f\left( x \right) = {\left( {{x^2} + 2x + 1} \right)^{\dfrac{1}{2}}} - 1 = {\left( {{{\left( {x + 1} \right)}^2}} \right)^{\dfrac{1}{2}}} - 1 = x \Rightarrow f\left( x \right) = x\)

Do đó \(P = f\left( {f\left( {2018} \right)} \right) = f\left( {2018} \right) = 2018\).