Bài 100803

Question

Giải phương trình :
$\begin{array}{l}
1) & {\log _3}\left( {\frac{3}{x}} \right){\log _2}x - {\log _3}\frac{{{x^3}}}{{\sqrt 3 }} = \frac{1}{2} + {\log _2}\sqrt x & & & \left( 1 \right)\\
2) & {\log _5}\frac{{{x^3}}}{{\sqrt 5 }} = \frac{1}{2} + \frac{{{{\log }_5}x}}{{{{\log }_3}\frac{1}{{\sqrt x }}}} & & & & & \left( 2 \right)
\end{array}$

score 0 · Answer 1

$\begin{array}{l}
1) ĐK: x > 0\\
\left( 1 \right) \Leftrightarrow (\log_33-\log_3x)\log_2x-(\log_3x^3-\log_3\sqrt3)=\frac{1}{2}+\log_2x^\frac{1}{2}\\ \Leftrightarrow \left( {1 - {{\log }_3}x} \right)\frac{{{{\log }_3}x}}{{{{\log }_3}2}} - (3{\log _3}x - \frac{1}{2}) = \frac{1}{2} + \frac{1}{2}\frac{{{{\log }_3}x}}{{{{\log }_3}2}}\\
\,\,\,\,\,\,\, \Leftrightarrow {\log _3}x\left( {\frac{1}{2} - {{\log }_3}x - 3{{\log }_3}2} \right) = 0\\
\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
{\log _3}x = 0\\
{\log _3}x = \frac{1}{2} - {\log _3}8
\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \frac{{\sqrt 3 }}{8}
\end{array} \right.(TM)
\end{array}$
Vậy PT đã cho có nghệm $x=3\vee x=\frac{\sqrt3}{8}$

$2) $
Đk: $\left\{ \begin{array}{l} x>0\\ x\neq 1 \end{array} \right.$
$(2)\Leftrightarrow (\log_5x^3-\log_55^\frac{1}{2})=\frac{1}{2}+\frac{\log_5x}{-\frac{1}{2}.\frac{\log_5x}{\log_53}}$
$\Leftrightarrow 3\log_5x-\frac{1}{2}=\frac{1}{2}-2\log_53$
$\Leftrightarrow \log_5x^3+\log_53^2=1$
$\Leftrightarrow \log_5(x^3.9)=\log_55$
$\Leftrightarrow 9x^3=5\Rightarrow x=\sqrt[3]{{\frac{5}{9}}}(TM)$
ĐS : $x = \sqrt[3]{{\frac{5}{9}}}$

Bài 100803

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Bài 100803

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