Số nghiệm nguyên của bất phương trình \(2{\log _2}\left( {2x} \right) + {\log _{\frac{1}{4}}}{x^2} < 5\) là

Ta có \(f\left( x \right) + f\left( {1 - x} \right) = \frac{1}{2}{\log _2}\left( {\frac{{2x}}{{1 - x}}} \right) + \frac{1}{2}{\log _2}\left( {\frac{{2\left( {1 - x} \right)}}{{1 - \left( {1 - x} \right)}}} \right)\)\( = \frac{1}{2}{\log _2}\frac{{2x}}{{1 - x}} + \frac{1}{2}{\log _2}\frac{{2\left( {1 - x} \right)}}{x}\)

\( = \frac{1}{2}{\log _2}\left[ {\frac{{2x}}{{1 - x}} \cdot \frac{{2\left( {1 - x} \right)}}{x}} \right]\)\( = \frac{1}{2}{\log _2}4 = 1\).

Ta có \(S = \left[ {f\left( {\frac{1}{{2025}}} \right) + f\left( {\frac{{2024}}{{2025}}} \right)} \right] + \left[ {f\left( {\frac{2}{{2025}}} \right) + f\left( {\frac{{2023}}{{2025}}} \right)} \right] + ... + \left[ {f\left( {\frac{{1012}}{{2025}}} \right) + f\left( {\frac{{1013}}{{2025}}} \right)} \right] = 1012\).

Ta có \({4^x} + {4^{ - x}} = 7\)\( \Leftrightarrow {2^{2x}} + {2^{ - 2x}} = 7\)\( \Leftrightarrow {\left( {{2^x}} \right)^2} + {\left( {{2^{ - x}}} \right)^2} = 7\)\( \Leftrightarrow {\left( {{2^x} + {2^{ - x}}} \right)^2} - 2 = 7\)\( \Leftrightarrow {\left( {{2^x} + {2^{ - x}}} \right)^2} = 9\)

\( \Leftrightarrow {2^x} + {2^{ - x}} = 3\).

Vậy \(P = \frac{{5 + {2^x} + {2^{ - x}}}}{{8 - 4 \cdot {2^x} - 4 \cdot {2^{ - x}}}}\)\( = \frac{{5 + 3}}{{8 - 4 \cdot 3}} =  - 2\).

Trả lời: −2.