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