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a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)
\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)
hay \(\widehat{yOz}=40^0\)
Vậy: \(\widehat{yOz}=40^0\)
\(3n-2\inƯ\left(15\right)\) \(=\left\{1;-1;3;-3;5;-5;15;-15\right\}.\)
\(\Leftrightarrow n\in\left\{1;\dfrac{1}{3};\dfrac{5}{3};\dfrac{-1}{3};\dfrac{7}{3};-1;\dfrac{17}{3};\dfrac{-13}{3}\right\}.\)
Mà \(n\ne\dfrac{2}{3};n\in Z.\)
\(\Rightarrow n\in\left\{1;-1\right\}.\)
\(d,=\left(-1000\right).38.25\left(-2\right)=\left(-38000\right).\left(-50\right)=1900000\\ e,=29.\left(2021+2020-2021\right)+\left(120-49\right).2020\\ =29.2020+\left(120-49\right).2020=2020.\left(29+120-49\right)=2020.100=202000\\ f,=\left(-25\right)\left(2023-22-1\right)=\left(-25\right).2000=-50000\)
a: \(=\left(-\dfrac{3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{1}{3}=\dfrac{1}{3}\)
b: \(=\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\dfrac{2}{3}\)
e: \(=\dfrac{-5}{21}-\dfrac{16}{21}+1=0\)
g: \(=\dfrac{-4}{11}\cdot\dfrac{-11}{4}\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
h: \(=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7}{36}+\dfrac{32}{36}-\dfrac{24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
i: \(=\dfrac{4}{7}-\dfrac{5}{8}-\dfrac{3}{28}=\dfrac{32}{56}-\dfrac{35}{56}-\dfrac{6}{56}=\dfrac{-9}{56}\)