Đề bài
Cho \(x < 0\). Chứng minh rằng: \(\sqrt {{{ - 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} }}} = {{1 - {2^x}} \over {1 + {2^x}}}\)
Lời giải chi tiết
Ta có: \(1 + {1 \over 4}{\left( {{2^x} - {2^{ - x}}} \right)^2} \)
\(\begin{array}{l}
= 1 + \frac{1}{4}\left( {{2^{2x}} - {{2.2}^x}{{.2}^{ - x}} + {2^{ - 2x}}} \right)\\
= 1 + \frac{1}{4}\left( {{4^x} - 2 + {4^{ - x}}} \right)
\end{array}\)
\(= {1 \over 4}\left( {4 + {4^x} - 2 + {4^{ - x}}} \right)\)
\(= {1 \over 4}\left( {{4^x} + 2 + {4^{ - x}}} \right) \)
\( = \frac{1}{4}\left( {{2^{2x}} + {{2.2}^x}{{.2}^{ - x}} + {2^{ - 2x}}} \right)\)
\(= {1 \over 4}{\left( {{2^x} + {2^{ - x}}} \right)^2}\)
\( \Rightarrow \sqrt {1 + \frac{1}{4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} \) \( = \sqrt {\frac{1}{4}{{\left( {{2^x} + {2^{ - x}}} \right)}^2}} \) \( = \frac{1}{2}\left( {{2^x} + {2^{ - x}}} \right)\)
Do đó:
\(\eqalign{
& \sqrt {{{ - 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} - {2^{ - x}}} \right)}^2}} }}} \cr&= \sqrt {{{ - 1 + {1 \over 2}\left( {{2^x} + {2^{ - x}}} \right)} \over {1 + {1 \over 2}\left( {{2^x} + {2^{ - x}}} \right)}}} \cr& = \sqrt {\frac{{\frac{{ - 2 + {2^x} + {2^{ - x}}}}{2}}}{{\frac{{2 + {2^x} + {2^{ - x}}}}{2}}}} \cr&= \sqrt {{{{2^x} - 2 + {2^{ - x}}} \over {{2^x} + 2 + {2^{ - x}}}}} \cr
& = \sqrt {{{{2^x} - 2 + {1 \over {{2^x}}}} \over {{2^x} + 2 + {1 \over {{2^x}}}}}}\cr& = \sqrt {\frac{{\frac{{{4^x} - {{2.2}^x} + 1}}{{{2^x}}}}}{{\frac{{{4^x} + {{2.2}^x} + 1}}{{{2^x}}}}}} \cr&= \sqrt {{{{4^x} - {{2.2}^x} + 1} \over {{4^x} + {{2.2}^x} + 1}}} \cr&= \sqrt {{{{{\left( {{2^x} - 1} \right)}^2}} \over {{{\left( {{2^x} + 1} \right)}^2}}}} = \frac{{\left| {{2^x} - 1} \right|}}{{\left| {{2^x} + 1} \right|}}\cr&= {{1 - {2^x}} \over {1 + {2^x}}} \cr} \)
(vì với \(x < 0\) thì \({2^x} < 1\))
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