1) a) \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}\)
\(=\left(-0,75+\frac{1}{2}\right)\cdot\frac{3}{4}\)
\(=\left(-0,75+0,5\right)\cdot0,75\)
\(=-0,25\cdot0,75\)
\(=-0,1875\)
Vậy: \(\left(-0,75+\frac{1}{2}\right):\frac{4}{3}=-0,1875\)
b) \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{32}{5}\)
\(=\frac{49}{20}-\frac{128}{20}\)
\(=-\frac{79}{20}=-3\frac{19}{20}\)
Vậy: \(\frac{7}{5}\cdot\frac{7}{4}-\frac{32}{5}=-3\frac{19}{20}\)
2) \(\frac{10}{3}\cdot x+\frac{67}{4}=-13,25\)
<=> \(\frac{10}{3}\cdot x+\frac{67}{4}=-\frac{53}{4}\)
<=> \(\frac{10}{3}\cdot x=-\left(\frac{53}{4}+\frac{67}{4}\right)\)
<=> \(\frac{10}{3}\cdot x=-30\)
<=> \(x=-30:\frac{10}{3}\)
<=> \(x=-30\cdot\frac{3}{10}\)
<=> \(x=-9\)
Vậy: \(x=-9\)
