Tính nhanh
\(58.\left(3\frac{1}{29}-2\frac{1}{28}\right):\frac{1}{3}\)
\(17\left(\frac{5}{17}-\frac{1}{34}+\frac{1}{2}\right):\frac{3}{4}\)
Giúp mình vs !!!!!!!!!!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
28 tháng 2 2022
a: \(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1-\dfrac{15}{17}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\left(\dfrac{12}{17}-1-\dfrac{15}{17}\right)+1\)
\(=\dfrac{-20}{17}+1=\dfrac{-3}{17}\)
b: \(B=\dfrac{-5}{3}\cdot16\dfrac{2}{7}-\dfrac{-5}{3}\cdot28\dfrac{2}{7}\)
\(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}-28-\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(-12\right)=20\)
c: \(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-25}{27}+\dfrac{1}{5}-1\)
\(=\dfrac{-125+27-135}{135}=\dfrac{-233}{135}\)
HH
21 tháng 7 2019
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
1 tháng 2 2020
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
Bài làm
a) \(58.\left(3\frac{1}{29}-2\frac{1}{28}\right):\frac{1}{3}\)
\(=58.\left(\frac{88}{29}-\frac{57}{28}\right).3\)
\(=58.\left(\frac{2464}{812}-\frac{1653}{812}\right).3\)
\(=58.\frac{811}{812}.3\)
\(=\frac{811}{14}.3\)
\(=\frac{2433}{3}\)
b) \(17.\left(\frac{5}{17}-\frac{1}{34}+\frac{1}{2}\right):\frac{3}{4}\)
\(=17.\left(\frac{10}{34}-\frac{1}{34}+\frac{17}{34}\right).\frac{4}{3}\)
\(=17.\frac{26}{34}.\frac{4}{3}\)
\(=\frac{26}{2}.\frac{4}{3}\)
\(=\frac{52}{3}\)
# Học tốt #