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Bài tập 81 trang 129 SGK Toán 12 NC

Bài tập 81 trang 129 SGK Toán 12 NC

Giải các bất phương trình sau:

\(\begin{array}{*{20}{l}}
{a){{\log }_5}(3x - 1) < 1}\\
{b){{\log }_{\frac{1}{3}}}(5x - 1) > 0}\\
{c){{\log }_{0,5}}({x^2} - 5x + 6) \ge  - 1}\\
{d){{\log }_3}\frac{{1 - 2x}}{x} \le 0.}
\end{array}\)

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Hướng dẫn giải chi tiết

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\log _5}(3x - 1) < 1\\
 \Leftrightarrow {\log _5}(3x - 1) < {\log _5}5
\end{array}\\
\begin{array}{l}
 \Leftrightarrow 0 < 3x - 1 < 5\\
 \Leftrightarrow \frac{1}{3} < x < 2
\end{array}
\end{array}\)

Vậy \(S = \left\{ {\frac{1}{3};2} \right\}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\log _{\frac{1}{3}}}(5x - 1) > 0\\
 \Leftrightarrow {\log _{\frac{1}{3}}}(5x - 1) > {\log _{\frac{1}{3}}}1
\end{array}\\
\begin{array}{l}
 \Leftrightarrow 0 < 5x - 1 < 1\\
 \Leftrightarrow \frac{1}{5} < x < \frac{2}{5}
\end{array}
\end{array}\)

Vậy \(S = \left\{ {\frac{1}{5};\frac{2}{5}} \right\}\)

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\log _{0,5}}({x^2} - 5x + 6) \ge  - 1\\
 \Leftrightarrow {\log _{0,5}}({x^2} - 5x + 6) \ge {\log _{0,5}}2
\end{array}\\
\begin{array}{l}
 \Leftrightarrow 0 < {x^2} - 5x + 6 \le 2\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{x^2} - 5x + 6 > 0}\\
{{x^2} - 5x + 4 \le 0}
\end{array}} \right.
\end{array}
\end{array}\)

\(\begin{array}{l}
 \Leftrightarrow 0 < {x^2} - 5x + 6 \le 2\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{x^2} - 5x + 6 > 0}\\
{{x^2} - 5x + 4 \le 0}
\end{array}} \right.
\end{array}\)

\(\begin{array}{l}
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x < 2{\mkern 1mu} ,x > 3}\\
{1 \le x \le 4}
\end{array}} \right.\\
 \Leftrightarrow 1 \le x < 2 \vee 3 < x \le 4
\end{array}\)

d)

\(\begin{array}{*{20}{l}}
{{{\log }_3}\frac{{1 - 2x}}{x} \le 0 \Leftrightarrow {{\log }_3}\frac{{1 - 2x}}{x} \le {{\log }_3}1}\\
{ \Leftrightarrow 0 < \frac{{1 - 2x}}{x} \le 1 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{\frac{{1 - 2x}}{x} > 0}\\
{\frac{{1 - 2x}}{x} - 1 \le 0}
\end{array}} \right.}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{0 < x < \frac{1}{2}}\\
{\frac{{1 - 3x}}{x} \le 0}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{0 < x < \frac{1}{2}}\\
{x \le 0;x \ge \frac{1}{3}}
\end{array}} \right.}\\
{ \Leftrightarrow \frac{1}{3} \le x < \frac{1}{2}}
\end{array}\)

Vậy \(S = \left[ {\frac{1}{3};\frac{1}{2}} \right)\)

-- Mod Toán 12 HỌC247

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