Bài tập 85 trang 130 SGK Toán 12 NC

Ta có:

\(\begin{array}{l}
1 + \frac{1}{4}{({2^x} - {2^{ - x}})^2} = \frac{1}{4}(4 + {4^x} - 2 + {4^{ - x}})\\
 = \frac{1}{4}({4^x} + 2 + {4^{ - x}}) = \frac{1}{4}{({2^x} + {2^{ - x}})^2}
\end{array}\)

Do đó: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt {\frac{{ - 1 + \sqrt {1 + \frac{1}{4}{{({2^x} - {2^{ - x}})}^2}} }}{{1 + \sqrt {1 + \frac{1}{4}{{({2^x} - {2^{ - x}})}^2}} }}} \\
 = \sqrt {\frac{{ - 1 + \frac{1}{2}({2^x} + {2^{ - x}})}}{{1 + \frac{1}{2}({2^x} + {2^{ - x}})}}} 
\end{array}\\
\begin{array}{l}
 = \sqrt {\frac{{{2^x} - 2 + {2^{ - x}}}}{{{2^x} + 2 + {2^{ - x}}}}}  = \sqrt {\frac{{{2^x} - 2 + \frac{1}{{{2^x}}}}}{{{2^x} + 2 + \frac{1}{{{2^x}}}}}} \\
 = \sqrt {\frac{{{4^x} - {{2.2}^x} + 1}}{{{4^x} + {{2.2}^x} + 1}}} 
\end{array}\\
{ = \sqrt {\frac{{{{\left( {{2^x} - 1} \right)}^2}}}{{{{\left( {{2^x} + 1} \right)}^2}}}}  = \frac{{1 - {2^x}}}{{1 + {2^x}}}}
\end{array}\)

(vì x < 0 thì 2^x < 1)

-- Mod Toán 12 HỌC247