Đáp án:
$\begin{array}{l}
a) - 10x\left( {x - \dfrac{1}{3}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - \dfrac{1}{3} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{1}{3}\\
b) - 2x\left( {x + \dfrac{1}{2}} \right).\left( {x - \dfrac{1}{7}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + \dfrac{1}{2} = 0\\
x - \dfrac{1}{7} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{1}{2}\\
x = \dfrac{1}{7}
\end{array} \right.\\
Vậy\,x = 0;x = - \dfrac{1}{2};x = \dfrac{1}{7}\\
B2)\\
\dfrac{x}{4} - \dfrac{1}{y} = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{x}{4} = \dfrac{1}{y} + \dfrac{1}{2}\\
\Leftrightarrow \dfrac{x}{4} = \dfrac{{y + 2}}{{2y}}\\
\Leftrightarrow 2xy = 4y + 8\\
\Leftrightarrow xy = 2y + 4\\
\Leftrightarrow xy - 2y = 4\\
\Leftrightarrow y\left( {x - 2} \right) = 4 = 1.4 = 2.2\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 4\\
x - 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 1\\
x - 2 = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 2\\
x - 2 = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - 4\\
x - 2 = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - 1\\
x - 2 = - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - 2\\
x - 2 = - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 4;x = 3\\
y = 1;x = 6\\
y = 2;x = 4\\
y = - 4;x = 1\\
y = - 1;x = - 2\\
y = - 2;x = 0
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {4;3} \right);\left( {1;6} \right);\left( {2;4} \right);\left( { - 4;1} \right);\left( { - 1; - 2} \right);\left( { - 2;0} \right)} \right\}\\
B3)\\
\dfrac{{ - 4}}{{11}}.\dfrac{{ - 3}}{{11}}.\dfrac{{ - 2}}{{11}}....\dfrac{3}{{11}}.\dfrac{4}{{11}}\\
= \dfrac{{ - 4}}{{11}}.\dfrac{{ - 3}}{{11}}.\dfrac{{ - 2}}{{11}}.\dfrac{{ - 1}}{{11}}.0.\dfrac{1}{{11}}.\dfrac{2}{{11}}\dfrac{3}{{11}}.\dfrac{4}{{11}}\\
= 0
\end{array}$
