Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\dfrac{{148 - x}}{{25}} + \dfrac{{169 - x}}{{23}} + \dfrac{{186 - x}}{{21}} + \dfrac{{199 - x}}{{19}} = 10\\
\Leftrightarrow \left( {\dfrac{{148 - x}}{{25}} - 1} \right) + \left( {\dfrac{{169 - x}}{{23}} - 2} \right) + \left( {\dfrac{{186 - x}}{{21}} - 3} \right) + \left( {\dfrac{{199 - x}}{{19}} - 4} \right) = 0\\
\Leftrightarrow \dfrac{{\left( {148 - x} \right) - 25}}{{25}} + \dfrac{{\left( {169 - x} \right) - 2.23}}{{23}} + \dfrac{{\left( {186 - x} \right) - 3.21}}{{21}} + \dfrac{{\left( {199 - x} \right) - 4.19}}{{19}} = 0\\
\Leftrightarrow \dfrac{{123 - x}}{{25}} + \dfrac{{123 - x}}{{23}} + \dfrac{{123 - x}}{{21}} + \dfrac{{123 - x}}{{19}} = 0\\
\Leftrightarrow \left( {123 - x} \right)\left( {\dfrac{1}{{25}} + \dfrac{1}{{23}} + \dfrac{1}{{21}} + \dfrac{1}{{19}}} \right) = 0\\
\Leftrightarrow 123 - x = 0\\
\Leftrightarrow x = 123\\
b,\\
\left| {4x - \dfrac{1}{3}} \right| = \dfrac{2}{5} \Leftrightarrow \left[ \begin{array}{l}
4x - \dfrac{1}{3} = \dfrac{2}{5}\\
4x - \dfrac{1}{3} = - \dfrac{2}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{{60}}\\
x = - \dfrac{1}{{60}}
\end{array} \right.\\
2,\\
a,\\
P = \dfrac{{2n - 3}}{{n + 4}} = \dfrac{{2.\left( {n + 4} \right) - 11}}{{n + 4}} = 2 - \dfrac{{11}}{{n + 4}}\\
P \in Z \Leftrightarrow \dfrac{{11}}{{n + 4}} \in Z \Leftrightarrow n + 4 \in \left\{ { \pm 1; \pm 11} \right\}\\
\Rightarrow n \in \left\{ { - 15; - 5; - 3;7} \right\}\\
b,\\
a - b = 2\left( {a + b} \right) = a:b\\
a - b = 2\left( {a + b} \right)\\
\Leftrightarrow 2\left( {a + b} \right) - \left( {a - b} \right) = 0\\
\Leftrightarrow a + 3b = 0\\
\Leftrightarrow a = - 3b\\
a - b = a:b\\
\Leftrightarrow \left( { - 3b} \right) - b = \left( { - 3b} \right):b\\
\Leftrightarrow - 4b = - 3\\
\Leftrightarrow b = \dfrac{3}{4} \Rightarrow a = - 3b = - \dfrac{9}{4}
\end{array}\)
