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a: \(-15-\left(-13+30\right)\)
\(=-15+13-30\)
\(=\left(-15-30\right)+13\)
=-45+13
=-32
b: \(225-150:\left(30+3^2\cdot5\right)\)
\(=225-150:\left(30+9\cdot5\right)\)
\(=225-150:75\)
=225-2
=223
c: \(18\cdot64+18\cdot36-1200\)
\(=18\left(64+36\right)-1200\)
\(=18\cdot100-12\cdot100\)
\(=6\cdot100=600\)
d: \(80-\left[130-\left(12-4\right)^2\right]\)
\(=80-130+\left(12-4\right)^2\)
\(=-50+8^2=64-50=14\)
a, 8-(3-7)
=8-(-4)
=4
b,(-5)-(9-12)
=(-5)-(-3)
=(-8)
c,7-(-9)-3
=(-2)-3
=-1
d.(-3)+8-11
=5-11
=6
\(\dfrac{5}{12}+\dfrac{3}{16}=\dfrac{5\cdot4+3\cdot3}{48}=\dfrac{29}{48}\)
\(\dfrac{4}{15}-\dfrac{2}{9}=\dfrac{4\cdot3-2\cdot5}{45}=\dfrac{2}{45}\)
a) \(\dfrac{5}{12}+\dfrac{3}{16}=\dfrac{20}{48}+\dfrac{9}{48}=\dfrac{29}{48}\)
b) \(\dfrac{4}{15}-\dfrac{2}{9}=\dfrac{12}{45}-\dfrac{10}{45}=\dfrac{2}{45}\)
a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)
\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)
b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
a) (–5).6 = – (|–5| . |6|) = – (5.6) = –30.
b) 9. (–3) = – (|9| . |–3|) = –(9.3) = –27.
c) (–10) .11 = – (|–10| . |11|) = –(10.11) = –110.
d) 150 . (–4) = – ( |150| . |–4|) = –(150.4) = –600.
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