\(3\cdot18\cdot12+4\cdot82\cdot9\)

\(=36\cdot18+36\cdot82\)

\(=36\cdot\left(18+82\right)\)

\(=36\cdot100\)

\(=3600\)

3.18.12+4.82.9

= 3.12.18+4.9.82

= 36.18+36.82

= 36.(18+82)

=36.100

3600

thanks

d)\(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4=\dfrac{1}{3^{50}}.\left(-3^{50}\right)-\dfrac{1}{6}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

e) \(=\left(4.\dfrac{3}{4}-\dfrac{1}{2}\right).\dfrac{6}{5}-17=\left(3-\dfrac{1}{2}\right).\dfrac{6}{5}-17=\dfrac{5}{2}.\dfrac{6}{5}-17=3-17=-14\)

\(=\dfrac{99\cdot44.44\cdot2\cdot13.20\cdot4}{3.3\cdot88.88\cdot2\cdot6.6\cdot8\cdot5}=\dfrac{99}{3.3}\cdot\dfrac{1}{2}\cdot2\cdot\dfrac{1}{2}\cdot\dfrac{1}{5}=30\cdot\dfrac{1}{5}=6\)

\(=\dfrac{3}{4}.\dfrac{-8}{5}+\dfrac{1}{4}=-\dfrac{6}{5}+\dfrac{1}{4}=-\dfrac{19}{20}\)

đề đâu bạn mik ko thấy

đề đâu rồi bạn ơi

\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right):\dfrac{2}{9}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right):\dfrac{2}{3}=\)

\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)\cdot\dfrac{3}{2}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\cdot\dfrac{3}{2}=\)

\(\left[\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\right]\cdot\dfrac{3}{2}=\)

\(\left[\dfrac{1}{2020}-\dfrac{2018}{2019}-\dfrac{2019}{2020}-\dfrac{1}{2019}\right]\cdot\dfrac{3}{2}=\)

\(0\)

\(\dfrac{1}{2}-\dfrac{1}{9}:2+1=\dfrac{13}{9}\)

\(\left(\dfrac{4}{13}-\dfrac{13}{5}\right):\dfrac{4}{5}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right):\dfrac{4}{5}\)

\(=\left(\dfrac{4}{13}-\dfrac{13}{5}\right)\cdot\dfrac{5}{4}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right)\cdot\dfrac{5}{4}\)

\(=\left[\dfrac{4}{13}-\dfrac{13}{5}+\dfrac{9}{13}+\dfrac{3}{5}\right]\cdot\dfrac{5}{4}\)

\(=\dfrac{-5}{4}\)

(-23)+13+(-17)+57

=[(-23)+13]+[(-17)+57]

=(-10)+40

=30

hok tốt

Bạn ưi..

Tính cách nhanh hay tính bth cx ra kq là 30 mak

Chắc do bạn tính sai đó

Bạn kiểm tra lại xem

^^

(-23)+13+(-17)+57

C1: -23+13+(-17)+57

=(-23+13)+(-17+57)

=-10+40

=30

C2: -23+13+(-17)+57

=[-23+(-17)]+(13+57)

=-40+70

=30

^_^