Kolejność działań:
- działania w nawiasach,
- potęgowanie, pierwiastkowanie,
- mnożenie,dzielenie,
- dodawanie, odejmowanie.
Skorzystano z praw potęgowania:
a⁰ = 1 dla a ≠ 0
a¹ = a dla każdego a
(a/b)⁻ⁿ = (b/a)ⁿ
(√a)² = a
16.
[tex]a) \ (5^{-1}+5)^{2} -(\dfrac{1}{25})^{-1}:5^{-2}-5^{-2} = (\dfrac{1}{5}+5)^{2}-(\dfrac{25}{1})^{1}:(\dfrac{1}{5})^{2}-(\dfrac{1}{5})^{2} =\\\\\\=(5\dfrac{1}{5})^{2}-25:\dfrac{1}{25}-\dfrac{1}{25}=(\dfrac{5\cdot5+1}{5})^{2}-25\cdot25-\dfrac{1}{25} = (\dfrac{26}{5})^{2}-625-\dfrac{1}{25}=\\\\\\=\dfrac{676}{25}-625\dfrac{1}{25}=27\dfrac{1}{25}-625\dfrac{1}{25} = \boxed{-598}[/tex]
[tex]b) \ (\dfrac{1}{3})^{-2}-(-\dfrac{3}{7})^{0} \cdot36\cdot(\sqrt{16})^{-1} =(\dfrac{3}{1})^{2}-1\cdot36\cdot4^{-1} = 3^{2}-36\cdot(\dfrac{1}{4})^{1} =9-36\cdot\dfrac{1}{4}=\\\\=9-9 = \boxed{0}[/tex]
[tex]c) \ \dfrac{(\dfrac{1}{2})^{-1}+(\dfrac{1}{5})^{-1}+(\dfrac{1}{27})^{-1}:(\dfrac{1}{3})^{-2}}{3\cdot(\sqrt{2}{3})^{-2}+(\dfrac{2}{9})^{-1}+(-\sqrt{3})^{0}}=\dfrac{2+5+27:3^{2}}{3\cdot(\sqrt{\frac{3}{2}})^{2}+\frac{9}{2}}+1}=\dfrac{7+27:9}{3\cdot\dfrac{3}{2}+\dfrac{9}{2}+1}=\\\\\\=\dfrac{7+3}{\dfrac{9}{2}+\dfrac{9}{2}+\dfrac{2}{2}} = \dfrac{10}{\dfrac{20}{2}}=\dfrac{10}{10} =\boxed{1}[/tex]
