Cho hàm số f ( x ) = ( x^1 + 1/2 log 4 x + 8^1/3 log x 2^2 + 1 )^1/2 − 1 với 0 < x # 1 . Tính giá trị biểu thức P = f ( f ( 2018 ) ) .

Ta có:

\[\begin{array}{*{20}{l}}{{x^{1 + \frac{1}{{2{{\log }_4}x}}}} = {x^{1 + \frac{1}{{{{\log }_2}x}}}} = {x^{1 + {{\log }_x}2}} = {x^{{{\log }_x}2x}} = 2x}\\{{8^{\frac{1}{{3{{\log }_{{x^2}}}2}}}} = {2^{3.\frac{1}{{3{{\log }_{{x^2}}}2}}}} = {2^{\frac{1}{{{{\log }_{{x^2}}}2}}}} = {2^{{{\log }_2}{x^2}}} = {x^2}}\end{array}\]

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

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