Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
3.\left( {x - \dfrac{1}{2}} \right) - 5.\left( {x + \dfrac{3}{5}} \right) =  - x + \dfrac{1}{5}\\
 \Leftrightarrow 3x - \dfrac{3}{2} - 5x - 5.\dfrac{3}{5} =  - x + \dfrac{1}{5}\\
 \Leftrightarrow 3x - \dfrac{3}{2} - 5x - 3 + x - \dfrac{1}{5} = 0\\
 \Leftrightarrow \left( {3x - 5x + x} \right) - \left( {\dfrac{3}{2} + 3 + \dfrac{1}{5}} \right) = 0\\
 \Leftrightarrow  - x - \dfrac{{47}}{{10}} = 0\\
 \Leftrightarrow x =  - \dfrac{{47}}{{10}}\\
b,\\
\left( {\dfrac{1}{4} - x} \right)\left( {x + \dfrac{2}{5}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{4} - x = 0\\
x + \dfrac{2}{5} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x =  - \dfrac{2}{5}
\end{array} \right.\\
c,\\
\dfrac{1}{4}.{x^2} - 3.x = 0\\
 \Leftrightarrow x.\left( {\dfrac{1}{4}x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
\dfrac{1}{4}x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
\dfrac{1}{4}x = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 12
\end{array} \right.\\
d,\\
\left| {2.x + 1} \right| + \dfrac{2}{3} = 2\\
 \Leftrightarrow \left| {2.x + 1} \right| = 2 - \dfrac{2}{3}\\
 \Leftrightarrow \left| {2x + 1} \right| = \dfrac{4}{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = \dfrac{4}{3}\\
2x + 1 =  - \dfrac{4}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{4}{3} - 1\\
2x =  - \dfrac{4}{3} - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{1}{3}\\
2x =  - \dfrac{7}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{6}\\
x =  - \dfrac{7}{6}
\end{array} \right.\\
e,\\
\left| {2.x - 3} \right| + \left| {y - 5} \right| = 0\\
\left. \begin{array}{l}
\left| {2.x - 3} \right| \ge 0,\,\,\,\forall x\\
\left| {y - 5} \right| \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow \left| {2x - 3} \right| + \left| {y - 5} \right| \ge 0,\,\,\,\forall x,y\\
 \Rightarrow \left\{ \begin{array}{l}
\left| {2x - 3} \right| = 0\\
\left| {y - 5} \right| = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - 3 = 0\\
y - 5 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = 5
\end{array} \right.\\
g,\\
{\left( {2x - 3} \right)^2} = 36\\
 \Leftrightarrow {\left( {2x - 3} \right)^2} = {6^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 6\\
2x - 3 =  - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 9\\
2x =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
h,\\
{7^{x + 2}} + {2.7^x} = 357\\
 \Leftrightarrow {7^x}{.7^2} + {2.7^x} = 357\\
 \Leftrightarrow {7^x}.\left( {{7^2} + 2} \right) = 357\\
 \Leftrightarrow {7^x}.51 = 357\\
 \Leftrightarrow {7^x} = 7\\
 \Leftrightarrow x = 1
\end{array}\)

$\begin{array}{l} a,\\ 3.\left( {x - \dfrac{1}{2}} \right) - 5.\left( {x + \dfrac{3}{5}} \right) = - x + \dfrac{1}{5}\\ \Leftrightarrow 3x - \dfrac{3}{2} - 5x - 5.\dfrac{3}{5} = - x + \dfrac{1}{5}\\ \Leftrightarrow 3x - \dfrac{3}{2} - 5x - 3 + x - \dfrac{1}{5} = 0\\ \Leftrightarrow \left( {3x - 5x + x} \right) - \left( {\dfrac{3}{2} + 3 + \dfrac{1}{5}} \right) = 0\\ \Leftrightarrow - x - \dfrac{{47}}{{10}} = 0\\ \Leftrightarrow x = - \dfrac{{47}}{{10}}\\ b,\\ \left( {\dfrac{1}{4} - x} \right)\left( {x + \dfrac{2}{5}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{4} - x = 0\\ x + \dfrac{2}{5} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{1}{4}\\ x = - \dfrac{2}{5} \end{array} \right.\\ c,\\ \dfrac{1}{4}.{x^2} - 3.x = 0\\ \Leftrightarrow x.\left( {\dfrac{1}{4}x - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ \dfrac{1}{4}x - 3 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ \dfrac{1}{4}x = 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 12 \end{array} \right.\\ d,\\ \left| {2.x + 1} \right| + \dfrac{2}{3} = 2\\ \Leftrightarrow \left| {2.x + 1} \right| = 2 - \dfrac{2}{3}\\ \Leftrightarrow \left| {2x + 1} \right| = \dfrac{4}{3}\\ \Leftrightarrow \left[ \begin{array}{l} 2x + 1 = \dfrac{4}{3}\\ 2x + 1 = - \dfrac{4}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = \dfrac{4}{3} - 1\\ 2x = - \dfrac{4}{3} - 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 2x = \dfrac{1}{3}\\ 2x = - \dfrac{7}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{1}{6}\\ x = - \dfrac{7}{6} \end{array} \right.\\ e,\\ \left| {2.x - 3} \right| + \left| {y - 5} \right| = 0\\ \left. \begin{array}{l} \left| {2.x - 3} \right| \ge 0,\,\,\,\forall x\\ \left| {y - 5} \right| \ge 0,\,\,\,\forall y \end{array} \right\} \Rightarrow \left| {2x - 3} \right| + \left| {y - 5} \right| \ge 0,\,\,\,\forall x,y\\ \Rightarrow \left\{ \begin{array}{l} \left| {2x - 3} \right| = 0\\ \left| {y - 5} \right| = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2x - 3 = 0\\ y - 5 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{3}{2}\\ y = 5 \end{array} \right.\\ g,\\ {\left( {2x - 3} \right)^2} = 36\\ \Leftrightarrow {\left( {2x - 3} \right)^2} = {6^2}\\ \Leftrightarrow \left[ \begin{array}{l} 2x - 3 = 6\\ 2x - 3 = - 6 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = 9\\ 2x = - 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{9}{2}\\ x = - \dfrac{3}{2} \end{array} \right.\\ h,\\ {7^{x + 2}} + {2.7^x} = 357\\ \Leftrightarrow {7^x}{.7^2} + {2.7^x} = 357\\ \Leftrightarrow {7^x}.\left( {{7^2} + 2} \right) = 357\\ \Leftrightarrow {7^x}.51 = 357\\ \Leftrightarrow {7^x} = 7\\ \Leftrightarrow x = 1 \end{array}$