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Câu 3:
a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)
\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)
b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)
\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)
Câu 4:
a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)
b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)
\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)
\(=-27\cdot100-23\cdot100\)
\(=100\left(-27-23\right)\)
\(=-50\cdot100=-5000\)
c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)
d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)
\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)
\(=-1+1+\dfrac{5}{6}\)
\(=\dfrac{5}{6}\)
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a: \(C=\dfrac{-1}{2}:\dfrac{-2}{3}:\dfrac{-3}{4}\)
\(=\dfrac{-1}{2}\cdot\dfrac{3}{-2}\cdot\dfrac{4}{-3}\)
\(=-\dfrac{4}{4}=-1\)
b: \(D=\dfrac{-6}{5}:\dfrac{-5}{4}:\dfrac{-4}{3}\)
\(=-\dfrac{6}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}\)
\(=\dfrac{-18}{25}\)
c: \(\dfrac{4}{20}+\dfrac{16}{42}+\dfrac{6}{15}+\dfrac{-3}{5}+\dfrac{2}{21}+\dfrac{-10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{4}{20}+\dfrac{-3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{16}{42}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}-\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{8}{21}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)
\(=\dfrac{3}{20}\)
d: \(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)
\(=\dfrac{21}{23}+\dfrac{125}{93}+\dfrac{-21}{21}+\dfrac{-125}{143}\)
\(=\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
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a) 36 ∉ BC(6; 21);
b) 3 ∈ ƯC(30; 42);
c) 30 ∉ BC(5; 12; 15);
d) 4 ∉ ƯC(16; 20; 30)
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\(a.=5-3+12-4-16\)
\(=-2\)
\(b.=-6-\left(-12\right)+7-10\)
\(=6+7-10\)
\(=3\)