Đáp án:
b. \(M = \dfrac{1}{{3x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
M = \dfrac{{\sqrt {9{x^2} - 6x + 1} }}{{9{x^2} - 1}}\\
a.DK:\left\{ \begin{array}{l}
9{x^2} - 6x + 1 \ge 0\\
9{x^2} - 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{\left( {3x - 1} \right)^2} \ge 0\left( {ld} \right)\\
\left( {3x - 1} \right)\left( {3x + 1} \right) \ne 0
\end{array} \right.\\
\to x \ne \pm \dfrac{1}{3}\\
b.M = \dfrac{{\sqrt {9{x^2} - 6x + 1} }}{{9{x^2} - 1}}\\
= \dfrac{{\sqrt {{{\left( {3x - 1} \right)}^2}} }}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{3x - 1}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}\\
= \dfrac{1}{{3x + 1}}\\
c.M = \dfrac{1}{4}\\
\to \dfrac{1}{{3x + 1}} = \dfrac{1}{4}\\
\to 3x + 1 = 4\\
\to 3x = 3\\
\to x = 1\\
d.M < 0\\
\to \dfrac{1}{{3x + 1}} < 0\\
\to 3x + 1 < 0\\
\to x < - \dfrac{1}{3}
\end{array}\)
