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PD
24 tháng 4 2018
CÂU 1:\(\frac{2}{35}:\left(\frac{3}{5}+\frac{3}{7}\right)\)
\(=\frac{2}{35}:\frac{36}{35}\)
\(=\frac{1}{18}\)
PD
24 tháng 4 2018
CÂU 2:\(\frac{1}{2}\cdot\frac{25}{6}-\left(\frac{4}{3}+\frac{5}{3}\right)\)
\(=\frac{25}{12}-3\)
\(=-\frac{11}{12}\)
10 tháng 5 2021
Giải:
a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\)
=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\)
=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\)
=\(\dfrac{7}{5}+\dfrac{-1}{4}\)
=\(\dfrac{23}{20}\)
b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\)
=\(\left[0+0\right]:\dfrac{2017}{2018}\)
=0\(:\dfrac{2017}{2018}\)
=0
10 tháng 5 2021
c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10
15 tháng 5 2022
1: =>x:7/3=-3/2
hay \(x=-\dfrac{3}{2}\cdot\dfrac{7}{3}=-\dfrac{7}{2}\)
2: \(x=\dfrac{11}{15}:\dfrac{2}{5}=\dfrac{11}{15}\cdot\dfrac{5}{2}=\dfrac{55}{30}=\dfrac{11}{6}\)
3: \(x=-\dfrac{6}{5}:\dfrac{3}{10}=\dfrac{-6}{5}\cdot\dfrac{10}{3}=\dfrac{-60}{15}=-4\)
4: \(x=\dfrac{1}{3}+\dfrac{3}{20}=\dfrac{20+9}{60}=\dfrac{29}{60}\)
6: \(\Leftrightarrow x\cdot\dfrac{13}{20}=\dfrac{5}{4}\)
hay \(x=\dfrac{5}{4}:\dfrac{13}{20}=\dfrac{5}{4}\cdot\dfrac{20}{13}=\dfrac{100}{52}=\dfrac{25}{13}\)
MC
21 tháng 7 2018
\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)
\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)
\(=\dfrac{231}{54}:\dfrac{7}{12}\)
\(=\dfrac{198}{27}\)
MC
21 tháng 7 2018
\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)
\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)
\(=\dfrac{0,8.0,64}{0,6}\)
\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)
LH
8 tháng 5 2017
\(B=\dfrac{2}{7}-\dfrac{6}{7}:3-3:\dfrac{2}{3}\)
\(B=\dfrac{2}{7}-\dfrac{2}{7}-\dfrac{9}{2}=0-\dfrac{9}{2}=-\dfrac{9}{2}\)
Vậy B \(=-\dfrac{9}{2}\)
\(C=1,2.\dfrac{25}{12}-\left(\dfrac{1}{2}+\dfrac{1}{3}\right):-\dfrac{5}{6}\)
\(C=\dfrac{6}{5}.\dfrac{25}{12}-\dfrac{5}{6}.\dfrac{-6}{5}\)
\(C=\dfrac{5}{2}-\left(-1\right)=\dfrac{7}{2}\)
Vậy \(C=\dfrac{7}{2}\)
1,2 . 25/6 - (4/3 + 5/3)
=12/10 . 25/6 - 3
=5 - 3
=2
chúc bạn hk giỏi nha
1,5 × 25/6- (4/3 + 5/3)
=>1,5 × 25/6 -3
=>25/4 - 3
=13/4
K MK NHA. CHÚC BẠN HỌC GIỎI