Bài tập 7 trang 212 SGK Toán 12 NC

Bài tập 7 trang 212 SGK Toán 12 NC

a) Ta có:

\(\begin{array}{l}
{\log _7}\frac{{a + b}}{3} = \frac{1}{2}({\log _7}a + {\log _7}b)\\
 \Leftrightarrow 2{\log _7}\frac{{a + b}}{3} = {\log _7}(ab)\\
 \Leftrightarrow {(\frac{{a + b}}{3})^2} = ab\\
 \Leftrightarrow {a^2} + 2ab + {b^2} = 9ab\\
 \Leftrightarrow {a^2} + {b^2} = 7ab\,\,(dpcm)
\end{array}\)

b) Ta có:

\(\begin{array}{l}
{\log _{a\sqrt b }}\frac{{\sqrt[3]{a}}}{{\sqrt {{b^3}} }} = \frac{{{{\log }_a}\frac{{\sqrt[3]{a}}}{{\sqrt {{b^3}} }}}}{{{{\log }_a}a\sqrt b }}\\
 = \frac{{{{\log }_a}\sqrt[3]{a} - {{\log }_a}\sqrt {{b^3}} }}{{{{\log }_a}a + {{\log }_a}\sqrt b }}\\
 = \frac{{\frac{1}{3} - \frac{3}{2}{{\log }_a}b}}{{1 + \frac{1}{2}{{\log }_a}b}} = \frac{{\frac{1}{3} - \frac{3}{2}\sqrt 3 }}{{1 + \frac{1}{2}\sqrt 3 }}\\
 = \frac{{2 - 9\sqrt 3 }}{{6 + 3\sqrt 3 }}
\end{array}\)

-- Mod Toán 12