Đáp án:
13) \(\left[ \begin{array}{l}
x = 7\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
13)\dfrac{1}{{{3^{\sqrt[3]{{{x^2} + 15}}}}}} = \dfrac{1}{{{9^{\sqrt[3]{{x + 1}}}}}}\\
\to \dfrac{1}{{{3^{\sqrt[3]{{{x^2} + 15}}}}}} = \dfrac{1}{{{{\left( {{3^2}} \right)}^{\sqrt[3]{{x + 1}}}}}}\\
\to \dfrac{1}{{{3^{\sqrt[3]{{{x^2} + 15}}}}}} = \dfrac{1}{{{3^{2.\sqrt[3]{{\left( {x + 1} \right)}}}}}}\\
\to \sqrt[3]{{{x^2} + 15}} = \sqrt[3]{{8x + 8}}\\
\to {x^2} + 15 = 8x + 8\\
\to \left[ \begin{array}{l}
x = 7\\
x = 1
\end{array} \right.\\
16){\left( {3\sqrt {{{3.3}^{\dfrac{1}{2}}}} } \right)^x} = {\left( {{3^{ - 4}}} \right)^{2x - 3}}\\
\to {\left( {3\sqrt {{3^{\dfrac{3}{2}}}} } \right)^x} = {\left( {{3^{ - 4}}} \right)^{2x - 3}}\\
\to {\left( {{{3.3}^{\dfrac{3}{2}:2}}} \right)^x} = {\left( {{3^{ - 4}}} \right)^{2x - 3}}\\
\to {\left( {{3^{\dfrac{7}{4}}}} \right)^x} = {\left( {{3^{ - 4}}} \right)^{2x - 3}}\\
\to {\left( 3 \right)^{\dfrac{7}{4}x}} = {3^{ - 8x + 12}}\\
\to \dfrac{7}{4}x = - 8x + 12\\
\to x = \dfrac{{16}}{{13}}
\end{array}\)
