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Câu hỏi: Biểu diễn các số sau đây theo a = ln2, b = ln5:
\(\ln 500;\ln {{16} \over {25}};\ln6,25\)
\(\ln{1 \over 2} + \ln {2 \over 3} + ... + \ln {{98} \over {99}} + \ln {{99} \over {100}}\).
\(\ln 500;\ln {{16} \over {25}};\ln6,25\)
\(\ln{1 \over 2} + \ln {2 \over 3} + ... + \ln {{98} \over {99}} + \ln {{99} \over {100}}\).
Lời giải chi tiết
+ ln500 = ln(53. 22) = ln53+ln22
= 3ln5+2ln2=3b+2a
+ ln(16/25)=ln16-ln25=ln24-ln(52)
=4ln2-2ln5=4a-2b
+ ln6,25=ln(625/100)=ln625-ln100
=ln(54)-ln(52. 22)=4ln5 - 2ln5-2ln2
=2ln5-2ln2=2b-2a
\(\ln{1 \over 2} + \ln {2 \over 3} + ... + \ln {{98} \over {99}} + \ln {{99} \over {100}} \)
\(= \ln 1 - \ln 2 + \ln 2 - \ln 3 + \) \(... + \ln99 - \ln100\)
\(= - \ln100 = - \ln\left( {{2^2}{{. 5}^2}} \right) \)
\(= - 2\ln 2 - 2\ln 5 = - 2a - 2b\).
Cách khác:
Ta có:
$\begin{array}{l}
\ln \frac{1}{2} + \ln \frac{2}{3} + ... + \ln \frac{{98}}{{99}} + \ln \frac{{99}}{{100}}\\
= \ln \frac{1}{2}.\frac{2}{3}...\frac{{98}}{{99}}.\frac{{99}}{{100}} = \ln \frac{1}{{100}} = - \ln 100\\
= - \ln \left( {{5^2}{{.2}^2}} \right) = - 2\ln 5 - 2\ln 2 = - 2b - 2a
\end{array}$
+ ln500 = ln(53. 22) = ln53+ln22
= 3ln5+2ln2=3b+2a
+ ln(16/25)=ln16-ln25=ln24-ln(52)
=4ln2-2ln5=4a-2b
+ ln6,25=ln(625/100)=ln625-ln100
=ln(54)-ln(52. 22)=4ln5 - 2ln5-2ln2
=2ln5-2ln2=2b-2a
\(\ln{1 \over 2} + \ln {2 \over 3} + ... + \ln {{98} \over {99}} + \ln {{99} \over {100}} \)
\(= \ln 1 - \ln 2 + \ln 2 - \ln 3 + \) \(... + \ln99 - \ln100\)
\(= - \ln100 = - \ln\left( {{2^2}{{. 5}^2}} \right) \)
\(= - 2\ln 2 - 2\ln 5 = - 2a - 2b\).
Cách khác:
Ta có:
$\begin{array}{l}
\ln \frac{1}{2} + \ln \frac{2}{3} + ... + \ln \frac{{98}}{{99}} + \ln \frac{{99}}{{100}}\\
= \ln \frac{1}{2}.\frac{2}{3}...\frac{{98}}{{99}}.\frac{{99}}{{100}} = \ln \frac{1}{{100}} = - \ln 100\\
= - \ln \left( {{5^2}{{.2}^2}} \right) = - 2\ln 5 - 2\ln 2 = - 2b - 2a
\end{array}$