2\(\times\)9\(\times\)3 / 9\(\times\)6\(\times\)13=
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)
\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)
\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)
a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)
\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)
\(A=\dfrac{-1}{10.2}\)
\(A=\dfrac{-1}{20}\)
b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)
\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)
\(B=\dfrac{2}{3.3}\)
\(B=\dfrac{2}{9}\)
c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)
\(C=\dfrac{2}{3}+\dfrac{2}{3}\)
\(C=\dfrac{4}{3}\)
d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)
\(D=\dfrac{-57}{57}=-1\)
e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)
\(E=\dfrac{27}{182}-\dfrac{27}{182}\)
\(E=0\)
a) \(\dfrac{2}{4}\times\dfrac{9}{5}=\dfrac{1}{2}\times\dfrac{9}{5}=\dfrac{1\times9}{2\times5}=\dfrac{9}{10}\)
b) \(\dfrac{13}{8}\times\dfrac{5}{15}=\dfrac{13}{8}\times\dfrac{1}{3}=\dfrac{13}{24}\)
c) \(\dfrac{3}{9}\times\dfrac{6}{12}=\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{3\times2}=\dfrac{1}{6}\)
a: =18/20=9/10
b: \(=\dfrac{13\cdot5}{8\cdot15}=\dfrac{13}{8}\cdot\dfrac{1}{3}=\dfrac{13}{24}\)
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{2}=\dfrac{1}{6}\)
\(A=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}\)
=1/3+1/2+2/5+1/3+2/7+1/4+1/9
=2789/1260
\(B=\dfrac{7}{3\cdot4}+\dfrac{9}{4\cdot5}+\dfrac{11}{5\cdot6}+\dfrac{13}{6\cdot7}+\dfrac{15}{7\cdot8}+\dfrac{17}{8\cdot9}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(=\dfrac{4}{9}+\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{2}{7}+\dfrac{1}{4}\)
=2789/1260
a, A = (\(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\)) - (\(\dfrac{79}{67}-\dfrac{28}{41}\))
= \(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
= \(\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
= \(\dfrac{1}{3}-1+1\)
= \(\dfrac{1}{3}\)
@Mai Tran