Bài 84: So sánh p và q, biết:
\(\eqalign{
& a)\,{\left( {{2 \over 3}} \right)^p} > {\left( {{3 \over 2}} \right)^{ – q}} \cr
& c)\,0,{25^p} < {\left( {{1 \over 2}} \right)^{2q}} \cr} \)
\(\eqalign{
& b)\,{\left( {{8 \over 3}} \right)^{ – p}} < {\left( {{3 \over 8}} \right)^q} \cr
& d)\,{\left( {{7 \over 2}} \right)^p} < {\left( {{2 \over 7}} \right)^{p – 2q}} \cr} \)
\(\eqalign{
& a)\,{\left( {{2 \over 3}} \right)^p} > {\left( {{3 \over 2}} \right)^{ – q}} \Leftrightarrow {\left( {{2 \over 3}} \right)^p} > {\left( {{2 \over 3}} \right)^q}\cr& \Leftrightarrow p < q\,\,\left( {\text{ vì }\,\,\,{2 \over 3} < 1} \right) \cr
& b)\,{\left( {{8 \over 3}} \right)^{ – p}} < {\left( {{3 \over 8}} \right)^q} \Leftrightarrow {\left( {{3 \over 8}} \right)^p} < {\left( {{3 \over 8}} \right)^q} \cr&\Leftrightarrow p > q\,\,\left( {\text{ vì }\,\,{3 \over 8} < 1} \right) \cr
& c)\,\,0,{25^p} < {\left( {{1 \over 2}} \right)^{2q}} \Leftrightarrow {\left( {{1 \over 4}} \right)^p} < {\left( {{1 \over 4}} \right)^q}\cr& \Leftrightarrow \,\,p > q\,\,\left( {\text{ vì }\,\,{1 \over 4} < 1} \right) \cr
& d)\,\,{\left( {{7 \over 2}} \right)^p} < {\left( {{2 \over 7}} \right)^{p – 2q}} \Leftrightarrow {\left( {{7 \over 2}} \right)^p} < {\left( {{7 \over 2}} \right)^{2q – p}} \cr&\Leftrightarrow p < 2q – p\,\,\left( {\text{ vì }\,\,{7 \over 2} > 1} \right) \cr
& \Leftrightarrow 2p < 2q \Leftrightarrow p < q \cr} \)
Bài 85: Cho \(x < 0\). Chứng minh rằng: \(\sqrt {{{ – 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} }}} = {{1 – {2^x}} \over {1 + {2^x}}}\)
Ta có: \(1 + {1 \over 4}{\left( {{2^x} – {2^{ – x}}} \right)^2} = {1 \over 4}\left( {4 + {4^x} – 2 + {4^{ – x}}} \right) \)
\(= {1 \over 4}\left( {{4^x} + 2 + {4^{ – x}}} \right) = {1 \over 4}{\left( {{2^x} + {2^{ – x}}} \right)^2}\)
Do đó:
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\(\eqalign{
& \sqrt {{{ – 1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} } \over {1 + \sqrt {1 + {1 \over 4}{{\left( {{2^x} – {2^{ – x}}} \right)}^2}} }}} \cr&= \sqrt {{{ – 1 + {1 \over 2}\left( {{2^x} + {2^{ – x}}} \right)} \over {1 + {1 \over 2}\left( {{2^x} + {2^{ – x}}} \right)}}} = \sqrt {{{{2^x} – 2 + {2^{ – x}}} \over {{2^x} + 2 + {2^{ – x}}}}} \cr
& = \sqrt {{{{2^x} – 2 + {1 \over {{2^x}}}} \over {{2^x} + 2 + {1 \over {{2^x}}}}}} = \sqrt {{{{4^x} – {{2.2}^x} + 1} \over {{4^x} + {{2.2}^x} + 1}}}\cr& = \sqrt {{{{{\left( {{2^x} – 1} \right)}^2}} \over {{{\left( {{2^x} + 1} \right)}^2}}}} = {{1 – {2^x}} \over {1 + {2^x}}} \cr} \)
(vì với \(x < 0\) thì \({2^x} < 1\))
Bài 86: Tính:
\(a)\,A = {9^{2{{\log }_3}4 + 4{{\log }_{81}}2}}\)
\(b)\,B = {\log _a}\left( {{{{a^2}.\root 3 \of a .\root 5 \of {{a^4}} } \over {\root 4 \of a }}} \right)\)
\(c)\,\,C = {\log _5}{\log _5}\root 5 \of {\root 5 \of {\root 5 \of {…\root 5 \of 5 } } } \)
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a) Áp dụng \({\log _{{a^\alpha }}}{b^\beta } = {\beta \over \alpha }{\log _a}b\) (với \(a > 0, b>0\) và \(a \ne 1\)) và \({a^{{{\log }_a}b}} = b\)
Ta có:
\(\eqalign{
& 2{\log _3}4 + 4{\log _{81}}2 = {4 \over 2}{\log _3}4 + 2{\log _9}2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\log _9}{4^4} + {\log _9}{2^2} = {\log _9}{2^{10}} \cr} \)
Do đó \(A = {9^{{{\log }_9}{2^{10}}}} = {2^{10}} = 1024\)
b) Ta có \({{{a^2}.\root 3 \of a .\root 5 \of {{a^4}} } \over {\root 4 \of a }} = {a^{2 + {1 \over 3} + {4 \over 5} – {1 \over 4}}} = {a^{{{173} \over {60}}}}\)
Do đó: \(B = {\log _a}{a^{{{173} \over {60}}}} = {{173} \over {60}}\)
c) Ta có \(\root 5 \of {\root 5 \of {\root 5 \of {…\root 5 \of 5 } } } = {5^{{{\left( {{1 \over 5}} \right)}^n}}} \Rightarrow {\log _5}\root 5 \of {\root 5 \of {\root 5 \of {…\root 5 \of 5 } } } \)
\(= {\left( {{1 \over 5}} \right)^n} = {5^{ – n}}\)
\( \Rightarrow C = – n\)
Bài 87: Chứng minh rằng \({\log _2}3 > {\log _3}4\)
Ta có \({\log _2}3 > {\log _3}4 \Leftrightarrow {1 \over {{{\log }_3}2}} > {\log _3}4 \Leftrightarrow {\log _3}2.{\log _3}4 < 1\) (vì \({\log _3}2 > 0\))
Áp dụng BĐT cô si cho hai số dương ta có:
\(\eqalign{
& \sqrt {{{\log }_3}2.{{\log }_3}4} < {1 \over 2}\left( {{{\log }_3}2 + {{\log }_3}4} \right) = {1 \over 2}{\log _3}8 \cr
&{1 \over 2}{\log _3}8 < {1 \over 2}{\log _3}9 = 1 \Rightarrow {\log _3}2.{\log _3}4 < 1\,\,\left( {dpcm} \right) \cr} \)
