Cho a>b>c>0. CMR: b/ (√a+b – √a-b) < c/ (√a+c - √a-c) Question Cho a>b>c>0. CMR: b/ (√a+b – √a-b) < c/ (√a+c - √a-c) in progress 0 Toán Alice 1 năm 2021-08-09T01:46:34+00:00 2021-08-09T01:46:34+00:00 1 Answers 4 views 0
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Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{b}{{\sqrt {a + b} – \sqrt {a – b} }} – \dfrac{c}{{\sqrt {a + c} – \sqrt {a – c} }}\\
= \dfrac{{b.\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}}{{\left( {\sqrt {a + b} – \sqrt {a – b} } \right)\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}} – \dfrac{{c\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}{{\left( {\sqrt {a + c} – \sqrt {a – c} } \right)\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}\\
= \dfrac{{b.\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}}{{\left( {a + b} \right) – \left( {a – b} \right)}} – \dfrac{{c\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}{{\left( {a + c} \right) – \left( {a – c} \right)}}\\
= \dfrac{{b.\left( {\sqrt {a + b} + \sqrt {a – b} } \right)}}{{2b}} – \dfrac{{c\left( {\sqrt {a + c} + \sqrt {a – c} } \right)}}{{2c}}\\
= \dfrac{{\sqrt {a + b} + \sqrt {a – b} }}{2} – \dfrac{{\sqrt {a + c} + \sqrt {a – c} }}{2}\\
= \dfrac{{\left( {\sqrt {a + b} – \sqrt {a + c} } \right) + \left( {\sqrt {a – b} – \sqrt {a – c} } \right)}}{2}\\
= \dfrac{{\dfrac{{\left( {\sqrt {a + b} – \sqrt {a + c} } \right)\left( {\sqrt {a + b} + \sqrt {a + c} } \right)}}{{\sqrt {a + b} + \sqrt {a + c} }} + \dfrac{{\left( {\sqrt {a – b} – \sqrt {a – c} } \right)\left( {\sqrt {a – b} + \sqrt {a – c} } \right)}}{{\sqrt {a – b} + \sqrt {a – c} }}}}{2}\\
= \dfrac{{\dfrac{{\left( {a + b} \right) – \left( {a + c} \right)}}{{\sqrt {a + b} + \sqrt {a + c} }} + \dfrac{{\left( {a – b} \right) – \left( {a – c} \right)}}{{\sqrt {a – b} + \sqrt {a – c} }}}}{2}\\
= \dfrac{1}{2}.\left[ {\dfrac{{b – c}}{{\sqrt {a + b} + \sqrt {a + c} }} + \dfrac{{ – b + c}}{{\sqrt {a – b} + \sqrt {a – c} }}} \right]\\
= \dfrac{1}{2}.\left( {b – c} \right).\left[ {\dfrac{1}{{\sqrt {a + b} + \sqrt {a + c} }} – \dfrac{1}{{\sqrt {a – b} + \sqrt {a – c} }}} \right]\\
\left. \begin{array}{l}
\sqrt {a + b} > \sqrt {a – b} \\
\sqrt {a + c} > \sqrt {a – c}
\end{array} \right\} \Rightarrow \sqrt {a + b} + \sqrt {a + c} > \sqrt {a – b} + \sqrt {a – c} \\
\Rightarrow \dfrac{1}{{\sqrt {a + b} + \sqrt {a + c} }} < \dfrac{1}{{\sqrt {a – b} + \sqrt {a – c} }}\\
a > b > c > 0 \Rightarrow \dfrac{1}{2}.\left( {b – c} \right).\left[ {\dfrac{1}{{\sqrt {a + b} + \sqrt {a + c} }} – \dfrac{1}{{\sqrt {a – b} + \sqrt {a – c} }}} \right] < 0\\
\Rightarrow \dfrac{b}{{\sqrt {a + b} – \sqrt {a – b} }} – \dfrac{c}{{\sqrt {a + c} – \sqrt {a – c} }} < 0\\
\Leftrightarrow \dfrac{b}{{\sqrt {a + b} – \sqrt {a – b} }} < \dfrac{c}{{\sqrt {a + c} – \sqrt {a – c} }}
\end{array}\)