Đáp án:
$\begin{array}{l}
2b)\dfrac{{4x + 1}}{{6x - 1}} < \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{4x + 1}}{{6x - 1}} - \dfrac{2}{3} < 0\\
\Leftrightarrow \dfrac{{3.\left( {4x + 1} \right) - 2.\left( {6x - 1} \right)}}{{3\left( {6x - 1} \right)}} < 0\\
\Leftrightarrow \dfrac{{12x + 3 - 12x + 2}}{{3\left( {6x - 1} \right)}} < 0\\
\Leftrightarrow \dfrac{5}{{3.\left( {6x - 1} \right)}} < 0\\
\Leftrightarrow 6x - 1 < 0\\
\Leftrightarrow x < \dfrac{1}{6}\\
Vậy\,x < \dfrac{1}{6}
\end{array}$
Bài 4:
a) Xét ΔABH và ΔAHM có:
+ góc BAH chung
+ góc AHB = góc AMH = 90 độ
=> ΔABH ~ ΔAHM (g-g)
$\begin{array}{l}
\Leftrightarrow \dfrac{{AB}}{{AH}} = \dfrac{{AH}}{{AM}}\\
\Leftrightarrow A{H^2} = AB.AM
\end{array}$
Tương tự ta cm được: ΔACH ~ ΔAHN (g-g)
$\begin{array}{l}
\Leftrightarrow \dfrac{{AC}}{{AH}} = \dfrac{{AH}}{{AN}}\\
\Leftrightarrow A{H^2} = AC.AN
\end{array}$
b)
$\begin{array}{l}
Do:A{H^2} = AB.AM = AC.AN\\
\Leftrightarrow \dfrac{{AB}}{{AC}} = \dfrac{{AN}}{{AM}}\\
Xet:\Delta AMN;\Delta ACB:\\
+ \dfrac{{AB}}{{AC}} = \dfrac{{AN}}{{AM}}\\
+ \widehat {BAC}\,chung\\
\Leftrightarrow \Delta AMN \sim \Delta ACB\left( {c - g - c} \right)
\end{array}$
