Giải thích các bước giải:
$a, \frac{3}{4} - (x - \frac{2}{5})= \frac{11}{12}$
$→\frac{3}{4}- x + \frac{2}{5} = \frac{11}{12}$
$→ \frac{23}{20} - x = \frac{11}{12}$
$→ x = \frac{23}{20} - \frac{11}{12}$
$→ x = \frac{7}{30}$
$b,|x + \frac{1}{3}| + 1 = 4$
$→ |x + \frac{1}{3}| = 3$
$→\left[ \begin{array}{l}x + \frac{1}{3} = 3\\x +\frac{1}{3} = - 3\end{array} \right.→\left[ \begin{array}{l}x=\frac{8}{3}\\x=\frac{-10}{3}\end{array} \right.$
$c, 27^{x} = 3^{x +2}$
$→ 3^{3x} = 3^{x +2}$
$→ 3x = x + 2$
$→ 3x - x = 2$
$→ 2x = 2$
$→ x = 1$
$d, \frac{x}{3} = \frac{y}{7}$
$Theo$ $tính$ $chất$ $dãy$ $tỉ$ $số$ $bằng$ $nhau$ $ta$ $có:$
$\frac{x}{3} = \frac{y}{7} = \frac{x-y}{3 - 7} = \frac{16}{-4} = - 4$
$→\left \{ {{x = - 4.3 = - 12} \atop {y = - 4.7 = - 28}} \right.$
$e, 4x = 5y → \frac{x}{5} = \frac{y}{4}$
$Theo$ $tính$ $chất$ $dãy$ $tỉ$ $số$ $bằng$ $nhau$ $ta$ $có:$
$\frac{x}{5} = \frac{y}{4} = \frac{x + y}{5 + 4} = \frac{36}{9} = 4$
$→\left \{ {{x = 4.5 = 20} \atop {y = 4.4 = 16}} \right.$
