Bài 44 trang 97 toán 12 nâng cao năm 2024
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Câu hỏi: Chứng minh: \({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left({\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left({\sqrt 2 - 1} \right) = 0\) 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: $\begin{array}{l} \ln {\left( {3 + 2\sqrt 2 } \right){\frac{7}{{16}}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 - 1} \right){\frac{{25}}{8}}}\\ \= \ln {\left( {{{\left( {\sqrt 2 + 1} \right)}2}} \right){\frac{7}{{16}}}} - \ln {\left( {\sqrt 2 + 1} \right)4} - \ln {\left( {\sqrt 2 + 1} \right){ - \frac{{25}}{8}}}\\ \= \ln {\left( {\sqrt 2 + 1} \right){\frac{7}{8}}} - \ln {\left( {\sqrt 2 + 1} \right)^4} - \ln {\left( {\sqrt 2 + 1} \right){ - \frac{{25}}{8}}}\\ \= \ln {\left( {\sqrt 2 + 1} \right)^{\frac{7}{8} - 4 + \frac{{25}}{8}}} = \ln {\left( {\sqrt 2 + 1} \right)^0} = \ln 1 = 0 \end{array}$ Các chủ đề tương tự
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\({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left( {\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left( {\sqrt 2 - 1} \right) = 0\) 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, \({7 \over {16}}\ln \left( {3 + 2\sqrt 2 } \right) - 4\ln \left( {\sqrt 2 + 1} \right) - {{25} \over 8}\ln \left( {\sqrt 2 - 1} \right)\) \( = {7 \over {16}}\ln {\left( {\sqrt 2 + 1} \right)2} - 4\ln \left( {\sqrt 2 + 1} \right) - {{25} \over 8}\ln {(\sqrt 2 + 1){-1} }\) \( = \frac{7}{{16}}.2\ln \left( {\sqrt 2 + 1} \right) - 4\ln \left( {\sqrt 2 + 1} \right) - \frac{{25}}{8}.\left( { - \ln \left( {\sqrt 2 + 1} \right)} \right)\) \( = {7 \over 8}\ln \left( {\sqrt 2 + 1} \right) - 4\ln \left( {\sqrt 2 + 1} \right) + {{25} \over 8}\ln \left( {\sqrt 2 + 1} \right) = 0\) |
