Упражнение 702

Решение:

• $${\largeа)}\ 0{,}68\cdot\frac{3}{4}=0{,}17\cdot3=0{,}51$$

• $${\largeб)}\ 3{,}212:\frac{4}{5}=3{,}212\cdot\frac{5}{4}=0{,}803\cdot5=4{,}015$$

• $${\largeв)}\ \frac{5}{6}\cdot24{,}6=5\cdot4{,}1=20{,}5$$

• $${\largeг)}\ 0{,}121:\frac{11}{12}=0{,}121\cdot\frac{12}{11}=0{,}011\cdot12=0{,}132$$

• $${\largeд)}\ 43{,}75\cdot\frac{2}{35}=1{,}25\cdot2=2{,}5$$

• $${\largeе)}\ \frac{13}{21}\cdot8{,}4=13\cdot0{,}4=5{,}2$$

• $${\largeж)}\ 5{,}6:3\frac{1}{2}=5{,}6:\frac{7}{2}=5{,}6\cdot\frac{2}{7}=0{,}8\cdot2=1{,}6$$

• $${\largeз)}\ 10\frac{2}{3}\cdot6{,}3=\frac{32}{3}\cdot6{,}3=32\cdot2{,}1=67{,}2$$

• $${\largeи)}\ 2\frac{3}{20}\cdot4{,}2=\frac{43}{20}\cdot4{,}2=43\cdot0{,}21=9{,}03$$

• $${\largeк)}\ \frac{2{,}3^{\backslash3}}{1{,}5\phantom{^{\backslash3}}}+\frac{6{,}7^{\backslash1}}{4{,}5\phantom{^{\backslash3}}}=\frac{6{,}9}{4{,}5}+\frac{6{,}7}{4{,}5}=\frac{13{,}6}{4{,}5}=\frac{136}{45}=3\frac{1}{45}$$

• $${\largeл)}\ \frac{1{,}5^{\backslash3}}{3{,}2\phantom{^{\backslash3}}}+\frac{1{,}9^{\backslash1}}{9{,}6\phantom{^{\backslash1}}}=\frac{4{,}5}{9{,}6}+\frac{1{,}9}{9{,}6}=\frac{6{,}4}{9{,}6}=\frac{64}{96}=\frac{2}{3}$$

• $${\largeм)}\ \frac{7{,}4^{\backslash2}}{5{,}7\phantom{^{\backslash2}}}-\frac{9{,}1^{\backslash1}}{11{,}4\phantom{^{\backslash1}}}=\frac{14{,}8}{11{,}4}-\frac{9{,}1}{11{,}4}=\frac{5{,}7}{11{,}4}=\frac{57}{114}=\frac{1}{2}$$