Ta có: \(1 + {1 \over 4}{\left( {{2^x} - {2^{ - x}}} \right)^2} = {1 \over 4}\left( {4 + {4^x} - 2 + {4^{ - x}}} \right) = {1 \over 4}\left( {{4^x} + 2 + {4^{ - x}}} \right) = {1 \over 4}{\left( {{2^x} + {2^{ - x}}} \right)^2}\)Do đó:\(\eqalign{
& \sqrt {{{ - 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} }}} = \sqrt {{{ - 1 + {1 \over 2}\left( {{2^x} + {2^{ - x}}} \right)} \over {1 + {1 \over 2}\left( {{2^x} + {2^{ - x}}} \right)}}} = \sqrt {{{{2^x} - 2 + {2^{ - x}}} \over {{2^x} + 2 + {2^{ - x}}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{{2^x} - 2 + {1 \over {{2^x}}}} \over {{2^x} + 2 + {1 \over {{2^x}}}}}} = \sqrt {{{{4^x} - {{2.2}^x} + 1} \over {{4^x} + {{2.2}^x} + 1}}} = \sqrt {{{{{\left( {{2^x} - 1} \right)}^2}} \over {{{\left( {{2^x} + 1} \right)}^2}}}} = {{1 - {2^x}} \over {1 + {2^x}}} \cr} \) (vì với \(x < 0\) thì \({2^x} < 1\))
