Câu 1:
\(\sqrt{9}\in N\)
\(1\dfrac{1}{9}\notin I\)
\(0,212121...\in Q\)
\(-5\notin I\)
\(-5\Omega\in R\)
\(-5\dfrac{1}{5}\notin Z\)
Câu 2:
a: \(\left[\left(2020^0-\dfrac{4}{3}\right)^2+\left|-\dfrac{5}{3}\right|\right]-\dfrac{\sqrt{16}}{9}\)
\(=\left[\left(1-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\right]-\dfrac{4}{9}\)
\(=\left[\dfrac{1}{9}+\dfrac{5}{3}\right]-\dfrac{4}{9}=\dfrac{5}{3}-\dfrac{3}{9}=\dfrac{4}{3}\)
b: \(0,5-\dfrac{21}{5}+2\dfrac{1}{2}\)
=0,5-4,2+2,5
=3-4,2
=-1,2
c: \(4,3\cdot\dfrac{7}{5}-\dfrac{7}{5}\cdot\dfrac{23}{10}+2023^0\)
\(=\dfrac{7}{5}\left(4,3-2,3\right)+1\)
\(=\dfrac{7}{5}\cdot2+1=\dfrac{14}{5}+1=\dfrac{19}{5}\)
Câu 1:
\(\sqrt{9}\in N\)
\(1\dfrac{1}{9}\notin I\)
\(0,212121...\in Q\)
\(-5\notin I\)
\(-5\pi\in R\)
\(-5\dfrac{1}{5}\notin Z\)
Câu 2:
\(a,\left[\left(2020^0-\dfrac{4}{3}\right)^2+\left|\dfrac{-5}{3}\right|\right]-\dfrac{\sqrt{16}}{9}\)
\(=\left[\left(1-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\right]-\dfrac{4}{9}\)
\(=\left[\left(\dfrac{-1}{3}\right)^2+\dfrac{5}{3}\right]-\dfrac{4}{9}\)
\(=\left[\dfrac{1}{9}+\dfrac{5}{3}\right]-\dfrac{4}{9}\)
\(=\left[\dfrac{1}{9}+\dfrac{15}{9}\right]-\dfrac{4}{9}\)
\(=\dfrac{16}{9}-\dfrac{4}{9}\)
\(=\dfrac{12}{9}=\dfrac{4}{3}\)
\(b,0,5-\dfrac{21}{5}+2\dfrac{1}{2}\)
\(=\dfrac{1}{2}-\dfrac{21}{5}+\dfrac{5}{2}\)
\(=\dfrac{5}{10}-\dfrac{42}{10}+\dfrac{25}{10}\)
\(=\dfrac{-12}{10}=\dfrac{-6}{5}\)
\(c,4,3.\dfrac{7}{5}-\dfrac{7}{5}.\dfrac{23}{10}+2023^0\)
\(=\dfrac{7}{5}\left(4,3-\dfrac{23}{10}\right)+1\)
\(=1,4\left(4,3-2,3\right)+1\)
\(=1,4.2+1\)
\(=2,8+1\)
\(=3,8\)