Odpowiedź:
[tex]\boxed{a)~~\left(0,7+0,65\cdot \dfrac{1}{5}\right) \div 10=0,083}[/tex]
[tex]\boxed{b)~~\left(5-3\dfrac{2}{9} \right)\cdot (3,2-2,3)=1,6}[/tex]
[tex]\boxed{c)~~\dfrac{26-3\cdot 4}{4+3\cdot 7} \div 1\dfrac{2}{5} =0,4}[/tex]
[tex]\boxed{d)~~\dfrac{4,2-2,4}{0,9} \cdot \dfrac{\dfrac{1}{6} +\dfrac{3}{5} -0,2}{3,2\div 0,8} =\dfrac{17}{60} }[/tex]
Szczegółowe wyjaśnienie:
Pamiętamy o kolejności wykonywanych działań:
- najpierw wykonujemy działania w nawiasach
- potem potęgowanie lub pierwiastkowanie
- następnie mnożenie lub dzielenie
- na końcu dodawanie lub odejmowanie.
Obliczamy:
[tex]a)\\\\\\\left(0,7+0,65\cdot \dfrac{1}{5}\right) \div 10=\left(0,7+\dfrac{65}{100} \cdot \dfrac{1}{5}\right) \div 10=\left(\dfrac{7}{10} + \dfrac{13}{100}\right) \div 10=\dfrac{83}{100} \div 10=\dfrac{83}{100}\cdot \dfrac{1}{10} =\dfrac{83}{1000}=0,083[/tex]
[tex]b)\\\\\left(5-3\dfrac{2}{9} \right)\cdot (3,2-2,3)=\left(4\dfrac{9}{9} -3\dfrac{2}{9} \right)\cdot 0,9=1\dfrac{7}{9} \right)\cdot \dfrac{9}{10} =\dfrac{16}{9} \right)\cdot \dfrac{9}{10} =\dfrac{16}{10} =1\dfrac{6}{10} =1\dfrac{3}{5} =1,6[/tex]
[tex]c)\\\\\dfrac{26-3\cdot 4}{4+3\cdot 7} \div 1\dfrac{2}{5} =\dfrac{26-12}{4+21} \div \dfrac{7}{5}=\dfrac{14}{25} \div \dfrac{7}{5}=\dfrac{14}{25} \cdot \dfrac{5}{7}= \dfrac{2}{5}= \dfrac{4}{10}=0,4[/tex]
[tex]d)\\\\\dfrac{4,2-2,4}{0,9} \cdot \dfrac{\dfrac{1}{6} +\dfrac{3}{5} -0,2}{3,2\div 0,8} =\dfrac{1,8}{0,9} \cdot \dfrac{\dfrac{1}{6} +\dfrac{3}{5} -\dfrac{2}{10} }{4} =\dfrac{1,8}{0,9} \cdot \dfrac{\dfrac{5}{30} +\dfrac{18}{30} -\dfrac{6}{30} }{4} =\left(\dfrac{18}{10} \div \dfrac{9}{10} \right)\cdot \left(\dfrac{17}{30} \div \dfrac{4}{1} \right)=\left(\dfrac{18}{10} \cdot \dfrac{10}{9} \right)\cdot \left(\dfrac{17}{30} \cdot \dfrac{1}{4} \right)=2 \cdot \dfrac{17}{120} =\dfrac{17}{60}[/tex]
