Giải thích các bước giải:
Bài 1:
a) $\dfrac{7}{5} \times x + \dfrac{3}{7}=\dfrac{-4}{5}\\⇔ \dfrac{7}{5} \times x = \dfrac{-4}{5}-\dfrac{3}{7}\\⇔ \dfrac{7}{5} \times x = \dfrac{-43}{35}\\⇔ x = \dfrac{-43}{35}: \dfrac{7}{5}\\⇔x = \dfrac{-43}{49}$
b) $|x|+0,573=2\\⇔|x|=2-0,573\\⇔|x|=1,427\\⇔ x = \pm 1,427$
c) $\bigg|x+\dfrac{1}{3}\bigg|-4=-1\\⇔\bigg|x+\dfrac{1}{3}\bigg|=3\\⇔\left[ \begin{array}{l}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{array} \right.\\ ⇔\left[ \begin{array}{l}x=3 - \dfrac{1}{3}\\x=-3-\dfrac{1}{3}\end{array} \right.\\⇔ \left[ \begin{array}{l}x=\dfrac{8}{3}\\x=\dfrac{-10}{3}\end{array} \right.$
Vậy `x ∈ {8/3; (-10)/3}`
Bài 2:
a) $\dfrac{-3}{7} < \dfrac{a}{5}<\dfrac{1}{4}\\⇔ \dfrac{-60}{140}<\dfrac{28a}{140}<\dfrac{35}{140}\\⇔ -60 < 28a<35\\⇔ \dfrac{-15}{7} < a < 1,25\\ \Rightarrow a \in \{ -2; -1; 0;1\}$
b) $\dfrac{2}{5}<\dfrac{3}{a}<\dfrac{1}{2}\\ ⇔ \dfrac{6}{15}< \dfrac{6}{2a}<\dfrac{6}{12}\\⇔ 15 > 2a>12\\⇔ 7,5>a>6\\⇔ a = 7$
