Реши пример

Это продолжение текущей учебной задачи. На фото я читаю №8 как \(3\frac12\cdot\frac7{10}-\frac34\), а №12 — как \(\dfrac{0,8}{\frac14-\frac15}\).

1. \(2{,}4-1{,}56:1{,}3=2{,}4-1{,}2=\boxed{1{,}2}\)

2. \(7{,}2\cdot3{,}5:0{,}9=25{,}2:0{,}9=\boxed{28}\)

3. \(2{,}4:4{,}8=\boxed{0{,}5}\)

4. \((7{,}5-3{,}6):2{,}5=3{,}9:2{,}5=\boxed{1{,}56}\)

5. \(8{,}1:(0{,}7-2{,}4)=8{,}1:(-1{,}7)=\boxed{-\dfrac{81}{17}}\)

6. \(-6{,}6-5\cdot(-3{,}5)=-6{,}6+17{,}5=\boxed{10{,}9}\)

7. \(\dfrac14+\dfrac9{20}=\dfrac5{20}+\dfrac9{20}=\boxed{\dfrac7{10}}\)

8. \(3\dfrac12\cdot\dfrac7{10}-\dfrac34 =\dfrac72\cdot\dfrac7{10}-\dfrac34 =\dfrac{49}{20}-\dfrac{15}{20} =\boxed{1{,}7}\)

9. \(2\dfrac45-(-0{,}6)\cdot\dfrac23 =2{,}8-(-0{,}4)=\boxed{3{,}2}\)

10. \(15\left(\dfrac3{20}+\dfrac7{30}\right) =15\cdot\dfrac{23}{60} =\boxed{5{,}75}\)

11. \(\left(-\dfrac78+4\dfrac23\right)\cdot9{,}6 =\dfrac{91}{24}\cdot\dfrac{48}{5} =\boxed{36{,}4}\)

12. \(\dfrac{0{,}8}{\frac14-\frac15} =\dfrac{0{,}8}{\frac1{20}} =\boxed{16}\)