\(b,\dfrac{7}{3}-1\dfrac{3}{5}.x=-1\dfrac{2}{3}\)
\(\Rightarrow\dfrac{8}{5}x=\dfrac{7}{3}+\dfrac{5}{3}\)
\(\Rightarrow\dfrac{8}{5}x=\dfrac{12}{3}\)
\(\Rightarrow\dfrac{8}{5}x=4\)
\(\Rightarrow x=4:\dfrac{8}{5}\)
\(\Rightarrow x=\dfrac{5}{2}\)
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\(d,\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=\left(-3\right)\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}=-3+7\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}=4\)
\(\Rightarrow\dfrac{1}{2}x=4+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{16}{4}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{19}{4}\)
\(\Rightarrow x=\dfrac{19}{4}:\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{19}{4}.2\)
\(\Rightarrow x=\dfrac{19}{2}\)
Vậy ...
