Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
7,2x + \left( { - 3,7x} \right) - 2,7 = 7,8\\
\Leftrightarrow 7,2x - 3,7x = 7,8 + 2,7\\
\Leftrightarrow 3,5x = 10,5\\
\Leftrightarrow x = 10,5:3,5\\
\Leftrightarrow x = 3\\
*)\\
\dfrac{2}{3} + \dfrac{1}{3}:\left| x \right| = 0,8\\
\Leftrightarrow \dfrac{1}{3}:\left| x \right| = 0,8 - \dfrac{2}{3}\\
\Leftrightarrow \dfrac{1}{3}:\left| x \right| = \dfrac{4}{5} - \dfrac{2}{3}\\
\Leftrightarrow \dfrac{1}{3}:\left| x \right| = \dfrac{2}{{15}}\\
\Leftrightarrow \left| x \right| = \dfrac{1}{3}:\dfrac{2}{{15}}\\
\Leftrightarrow \left| x \right| = \dfrac{5}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = - \dfrac{5}{2}
\end{array} \right.\\
*)\\
\left| {2x + 1} \right| + \dfrac{2}{3} = 2\\
\Leftrightarrow \left| {2x + 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}
x = \dfrac{1}{3}\\
x = - \dfrac{7}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{6}\\
x = - \dfrac{7}{6}
\end{array} \right.\\
*)\\
\left| {2x - 3} \right| + \left| {y - 5} \right| = 0\\
\left. \begin{array}{l}
\left| {2x - 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.\\
*)\\
{\left( {2x - 3} \right)^2} = 36\\
\Leftrightarrow \left| {2x - 3} \right| = 6\\
\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.\\
*)\\
{7^{x + 2}} + {2.7^x} = 357\\
\Leftrightarrow {7^x}{.7^2} + {2.7^x} = 357\\
\Leftrightarrow {7^x}.49 + {2.7^x} = 357\\
\Leftrightarrow {51.7^x} = 357\\
\Leftrightarrow {7^x} = 357:51\\
\Leftrightarrow {7^x} = 7\\
\Leftrightarrow x = 1
\end{array}\)
