Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f,\\
\dfrac{{2\left( {x + 1} \right)}}{3} - 1 = \dfrac{3}{2} - \dfrac{{1 - 2x}}{4}\\
\Leftrightarrow \dfrac{2}{3}x + \dfrac{2}{3} - 1 = \dfrac{3}{2} - \left( {\dfrac{1}{4} - \dfrac{1}{2}x} \right)\\
\Leftrightarrow \dfrac{2}{3}x - \dfrac{1}{3} = \dfrac{3}{2} - \dfrac{1}{4} + \dfrac{1}{2}x\\
\Leftrightarrow \dfrac{2}{3}x - \dfrac{1}{2}x = \dfrac{3}{2} - \dfrac{1}{4} + \dfrac{1}{3}\\
\Leftrightarrow \dfrac{1}{6}x = \dfrac{{19}}{{12}}\\
\Leftrightarrow x = \dfrac{{19}}{2}\\
g,\\
\dfrac{{3\left( {x + 1} \right)}}{{10}} = \dfrac{{2.\left( {x - 1} \right)}}{3} + \dfrac{4}{5}\\
\Leftrightarrow \dfrac{{3\left( {x + 1} \right)}}{{10}} = \dfrac{{10.\left( {x - 1} \right) + 12}}{{15}}\\
\Leftrightarrow \dfrac{{3\left( {x + 1} \right)}}{{10}} = \dfrac{{10x + 2}}{{15}}\\
\Leftrightarrow 3.\left( {x + 1} \right).15 = 10.\left( {10x + 2} \right)\\
\Leftrightarrow 3.\left( {x + 1} \right).3 = 2.\left( {10x + 2} \right)\\
\Leftrightarrow 9\left( {x + 1} \right) = 2.\left( {10x + 2} \right)\\
\Leftrightarrow 9x + 9 = 20x + 4\\
\Leftrightarrow 9 - 4 = 20x - 9x\\
\Leftrightarrow 5 = 11x\\
\Leftrightarrow x = \dfrac{5}{{11}}\\
h,\\
{x^2} - x - 2 = 0\\
\Leftrightarrow \left( {{x^2} - 2x} \right) + \left( {x - 2} \right) = 0\\
\Leftrightarrow x\left( {x - 2} \right) + \left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.
\end{array}\)
