## Task 2 - 180712

Calculate

$$
\begin{aligned}
& a=1.25: \frac{13}{12} \cdot \frac{91}{60} \\
& b=2.225-\frac{5}{9}-\frac{5}{6} \\
& c=\frac{32}{15}: \frac{14}{15}+6+\left(\frac{45}{56}-0.375\right) \\
& d=c-\frac{b}{a}
\end{aligned}
$$

without using approximate values!
$$
\begin{gathered}
a=\frac{5}{4}: \frac{13}{12} \cdot \frac{91}{60}=\frac{5 \cdot 12 \cdot 91}{4 \cdot 13 \cdot 60}=\frac{7}{4} \\
b=\frac{89}{40}-\frac{5}{9}-\frac{5}{6}=\frac{801-200-300}{360}=\frac{301}{360} \\
c=\frac{32 \cdot 15}{15 \cdot 14}+6+\left(\frac{45}{56}-\frac{3}{8}\right)=\frac{17}{6}+6+\frac{45-21}{56}=\frac{61}{7} \\
d=\frac{61}{7}-\frac{\frac{301}{360}}{\frac{7}{4}}=\frac{61}{7}-\frac{301 \cdot 4}{360 \cdot 7}=\frac{61 \cdot 90}{7 \cdot 90}-\frac{301}{90 \cdot 7}=\frac{5490-301}{630}=\frac{5189}{630}
\end{gathered}
$$

Note: The expressions $a=1 \frac{3}{4}, a=1.75, c=8 \frac{5}{7}$, and $d=8 \frac{149}{630}$ also satisfy the problem statement.