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Câu hỏi: Chứng minh:
\({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left({\sqrt 2 - 1} \right) = 0\)
\({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left({\sqrt 2 - 1} \right) = 0\)
Lời giải chi tiết
Ta có:
\(\begin{array}{l}
3 + 2\sqrt 2 = 2 + 2\sqrt 2 + 1\\
= {\left({\sqrt 2 } \right)^2} + 2\sqrt 2 + {1^2} = {\left({\sqrt 2 + 1} \right)^2}\\
\left({\sqrt 2 - 1} \right)\left({\sqrt 2 + 1} \right) = 2 - 1 = 1\\
\Rightarrow \sqrt 2 - 1 = \frac{1}{{\sqrt 2 + 1}}=(\sqrt 2 + 1)^{-1}
\end{array}\)
Do đó,
\({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left({\sqrt 2 - 1} \right)\)
\(= {7 \over {16}}\ln {\left( {\sqrt 2 + 1} \right)^2} - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln {(\sqrt 2 + 1)^{-1} }\)
\(= \frac{7}{{16}}. 2\ln \left( {\sqrt 2 + 1} \right) - 4\ln \left({\sqrt 2 + 1} \right) - \frac{{25}}{8}.\left({ - \ln \left( {\sqrt 2 + 1} \right)} \right)\)
\(= {7 \over 8}\ln \left( {\sqrt 2 + 1} \right) - 4\ln \left({\sqrt 2 + 1} \right) + {{25} \over 8}\ln \left({\sqrt 2 + 1} \right) = 0\)
Cách trình bày khác:
$\begin{array}{l}
\ln {\left( {3 + 2\sqrt 2 } \right)^{\frac{7}{{16}}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 - 1} \right)^{\frac{{25}}{8}}}\\
= \ln {\left( {{{\left( {\sqrt 2 + 1} \right)}^2}} \right)^{\frac{7}{{16}}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 + 1} \right)^{ - \frac{{25}}{8}}}\\
= \ln {\left( {\sqrt 2 + 1} \right)^{\frac{7}{8}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 + 1} \right)^{ - \frac{{25}}{8}}}\\
= \ln {\left( {\sqrt 2 + 1} \right)^{\frac{7}{8} - 4 + \frac{{25}}{8}}} = \ln {\left( {\sqrt 2 + 1} \right)^0} = \ln 1 = 0
\end{array}$
Ta có:
\(\begin{array}{l}
3 + 2\sqrt 2 = 2 + 2\sqrt 2 + 1\\
= {\left({\sqrt 2 } \right)^2} + 2\sqrt 2 + {1^2} = {\left({\sqrt 2 + 1} \right)^2}\\
\left({\sqrt 2 - 1} \right)\left({\sqrt 2 + 1} \right) = 2 - 1 = 1\\
\Rightarrow \sqrt 2 - 1 = \frac{1}{{\sqrt 2 + 1}}=(\sqrt 2 + 1)^{-1}
\end{array}\)
Do đó,
\({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left({\sqrt 2 - 1} \right)\)
\(= {7 \over {16}}\ln {\left( {\sqrt 2 + 1} \right)^2} - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln {(\sqrt 2 + 1)^{-1} }\)
\(= \frac{7}{{16}}. 2\ln \left( {\sqrt 2 + 1} \right) - 4\ln \left({\sqrt 2 + 1} \right) - \frac{{25}}{8}.\left({ - \ln \left( {\sqrt 2 + 1} \right)} \right)\)
\(= {7 \over 8}\ln \left( {\sqrt 2 + 1} \right) - 4\ln \left({\sqrt 2 + 1} \right) + {{25} \over 8}\ln \left({\sqrt 2 + 1} \right) = 0\)
Cách trình bày khác:
$\begin{array}{l}
\ln {\left( {3 + 2\sqrt 2 } \right)^{\frac{7}{{16}}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 - 1} \right)^{\frac{{25}}{8}}}\\
= \ln {\left( {{{\left( {\sqrt 2 + 1} \right)}^2}} \right)^{\frac{7}{{16}}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 + 1} \right)^{ - \frac{{25}}{8}}}\\
= \ln {\left( {\sqrt 2 + 1} \right)^{\frac{7}{8}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 + 1} \right)^{ - \frac{{25}}{8}}}\\
= \ln {\left( {\sqrt 2 + 1} \right)^{\frac{7}{8} - 4 + \frac{{25}}{8}}} = \ln {\left( {\sqrt 2 + 1} \right)^0} = \ln 1 = 0
\end{array}$