П.4. Вычислите: 1) 11 · 1 17/55 − 15,3; 2) 3 4/9 · 4,5 · 7/35 · 8,75 : 2/5 : 9/32; 3) 19 · 2 55/76 − 39,4; 4) 6,8 · 5/13 · 2,7 − 3 3/25 : 5 2/5 · 1/175.

1. \(11\cdot1\frac{17}{55}-15{,}3=11\cdot\frac{72}{55}-\frac{153}{10}=\frac{792}{55}-\frac{153}{10}=\frac{144}{10}-\frac{153}{10}=-\frac9{10}=-0{,}9\).
2. \(3\frac49\cdot4{,}5\cdot\frac7{35}\cdot8{,}75:\frac25:\frac9{32}=\frac{31}{9}\cdot\frac92\cdot\frac15\cdot\frac{35}{4}\cdot\frac52\cdot\frac{32}{9}=\frac{4960}{9}\).
3. \(19\cdot2\frac{55}{76}-39{,}4=19\cdot\frac{207}{76}-\frac{197}{5}=\frac{207}{4}-\frac{197}{5}=\frac{1035-788}{20}=\frac{247}{20}=12{,}35\).
4. \(6{,}8\cdot\frac5{13}\cdot2{,}7-3\frac3{25}:5\frac25\cdot\frac1{175}=\frac{34}{5}\cdot\frac5{13}\cdot\frac{27}{10}-\frac{78}{25}:\frac{27}{5}\cdot\frac1{175}\).
   \(=\frac{918}{65}-\frac{78}{25}\cdot\frac5{27}\cdot\frac1{175}=\frac{918}{65}-\frac{26}{23625}=\frac{3334756}{23625}\approx141{,}0479\).

Ответ: 1) \(-0{,}9\); 2) \(\frac{4960}{9}\); 3) \(12{,}35\); 4) \(\frac{3334756}{23625}\approx141{,}0479\).