Đáp án:
h. \(x = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a. - \frac{2}{3}x = \frac{5}{9}\\
\to x = \frac{5}{9}:\left( { - \frac{2}{3}} \right)\\
\to x = - \frac{5}{6}\\
c.{x^2} = \frac{4}{{25}}\\
\to x = \pm \sqrt {\frac{4}{{25}}} \\
\to x = \pm \frac{2}{5}\\
d.{\left( {x + 1} \right)^3} = - \frac{1}{8}\\
\to x + 1 = \sqrt[3]{{ - \frac{1}{8}}}\\
\to x + 1 = - \frac{1}{2}\\
\to x = - \frac{3}{2}\\
g.\left( {x - 1} \right)\left( {x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 5
\end{array} \right.\\
b.\frac{1}{2}x + \frac{3}{2} - \frac{3}{2}x + 3 - 0\\
\to - x + \frac{9}{2} = 0\\
\to x = \frac{9}{2}\\
d.\frac{1}{2}{x^3} = \frac{4}{{27}}\\
\to {x^3} = \frac{8}{{27}}\\
\to x = \sqrt[3]{{\frac{8}{{27}}}}\\
\to x = \frac{2}{3}\\
h.\left( {x + 1} \right)\left( {{x^2} + 1} \right) = 0\\
\to x + 1 = 0\left( {do:{x^2} + 1 > 0\forall x \in R} \right)\\
\to x = - 1
\end{array}\)
