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 \[{2^x} < 1\]]