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1)A=(logab+logba+2)(1−logaba)−1(a,b,ab>0,≠1)=(logab+logba+2)(1−11+logab)−1=11+logab(logab+logba+2)logab−1=11+logab(log2ab+1+2logab)−1=11+logab(logab+1)2−1=logab
\begin{array}{l} 2)B = 1 + 2{\log _2}x + {\log _2}x\left( {{{\log }_2}x + 1} \right) + \log _2^2x\,\,\, = 3\log _2^2x + 3{\log _2}x + 1\\ 3)n,p > 0;\,\,n,p,np \ne 1,\,\,\,{\log _n}p \ge 0\,\,\,(1)\\ C = \sqrt {{{\log }_n}p + {{\log }_p}n + 2} \left( {{{\log }_n}p - \frac{{{{\log }_n}p}}{{1 + {{\log }_n}p}}} \right)\sqrt {{{\log }_n}p} \\ = \sqrt {{{\log }_n}p\left( {{{\log }_n}p + {{\log }_p}n + 2} \right)} \left( {\frac{{\log _n^2p}}{{1 + {{\log }_n}p}}} \right)\\ = \sqrt {{{\left( {{{\log }_n}p + 1} \right)}^2}} \left( {\frac{{\log _n^2p}}{{1 + {{\log }_n}p}}} \right)\\ = \left| {{{\log }_n}p + 1} \right|.\frac{{\log _n^2p}}{{1 + {{\log }_n}p}} = \log _n^2p\\ C = \log _n^2p\,\,\,\,\,vì\,\,\,{\log _n}p + 1 > 0\\ \end{array}
(theo điều kiện (1))
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Đăng bài 24-04-12 09:21 AM
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