Question

Задания 1 и 2. По действиям. ​

Answer 1

$1) 1,68:1,6-2,1=1,05-2,1=-1,05\\\\2)-1,05*(-1\frac{2}{3}) = \frac{105}{100}*\frac{5}{3} =\frac{35}{20}=\frac{7}{4} =1\frac{3}{4} \\\\3)1\frac{3}{4}-2\frac{1}{6}=-(2 + \frac{1}{6} -1-\frac{3}{4})=-(1+\frac{2-9}{12})=-(1-\frac{7}{12})=-\frac{5}{12} \\\\Otvet:-\frac{5}{12}\\\\\\2)3^{2}:3^{-1}-(\sqrt[3]{125})^{2} -5*5^{-2}+(\sqrt{13})^{0} =3^{2-(-1)}-(\sqrt[3]{5^{3}})^{2} -5^{1-(-2)} +1=\\\\=3^{3}-5^{2}-5^{3} +1=27-25-125+1=-122$