Bài 85. Cho \(x < 0\). Chứng minh rằng: \(\sqrt {{{ – 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} }}} = {{1 – {2^x}} \over {1 + {2^x}}}\)
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:
\(\eqalign{
& \sqrt {{{ – 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} }}} = \sqrt {{{ – 1 + {1 \over 2}\left( {{2^x} + {2^{ – x}}} \right)} \over {1 + {1 \over 2}\left( {{2^x} + {2^{ – x}}} \right)}}} = \sqrt {{{{2^x} – 2 + {2^{ – x}}} \over {{2^x} + 2 + {2^{ – x}}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{{2^x} – 2 + {1 \over {{2^x}}}} \over {{2^x} + 2 + {1 \over {{2^x}}}}}} = \sqrt {{{{4^x} – {{2.2}^x} + 1} \over {{4^x} + {{2.2}^x} + 1}}} = \sqrt {{{{{\left( {{2^x} – 1} \right)}^2}} \over {{{\left( {{2^x} + 1} \right)}^2}}}} = {{1 – {2^x}} \over {1 + {2^x}}} \cr} \)
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(vì với \(x < 0\) thì \({2^x} < 1\))