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NH
4
LP
1
ND
Nguyễn Đăng Nhân
CTVHS
21 tháng 2 2023
a) \(\dfrac{2}{5}\cdot\dfrac{1}{7}+\dfrac{2}{5}\cdot\dfrac{5}{7}+\dfrac{2}{5}\)
\(=\dfrac{2}{5}\left(\dfrac{1}{7}+\dfrac{5}{7}+1\right)\)
\(=\dfrac{2}{5}\cdot\dfrac{13}{7}=\dfrac{26}{35}\)
b) \(\dfrac{1}{5}+\dfrac{2}{8}+\dfrac{4}{5}+\dfrac{7}{8}-\dfrac{1}{8}\)
\(=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{2}{8}+\dfrac{7}{8}-\dfrac{1}{8}\right)\)
\(=1+1=2\)
c)\(\dfrac{24}{36}\cdot\dfrac{10}{12}\cdot36\)
\(=\dfrac{24\cdot10\cdot36}{36\cdot12}=\dfrac{12\cdot2\cdot10\cdot36}{12\cdot36}\)
\(=2\cdot10=20\)
MH
3
15 tháng 4 2022
\(\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{8}{9}\right)=1+1+1+1=4\)
K
15 tháng 4 2019
a, \(\left(\frac{4}{9}.\frac{3}{7}\right).\frac{7}{4}=\frac{4}{9}\left(\frac{3}{7}.\frac{7}{4}\right)=\frac{4}{9}.\frac{3}{4}=\frac{3}{9}=\frac{1}{3}\)
b, \(\left(\frac{6}{5}.\frac{4}{5}\right).\frac{25}{16}=\frac{6}{5}\left(\frac{4}{5}.\frac{25}{16}\right)=\frac{6}{5}.\frac{5}{4}=\frac{6}{4}=\frac{3}{2}\)
c, \(\left(\frac{7}{8}.\frac{16}{9}\right).\frac{3}{14}=\frac{7}{8}\left(\frac{16}{9}.\frac{3}{14}\right)=\frac{7}{8}.\frac{8}{21}=\frac{7}{21}=\frac{1}{3}\)
14 tháng 4 2019
( 4/9 x 3/7 ) x 7/4 = 4/9 x ( 3/7 x 7/4 ) =4/9 x 21/28 = 84/252
( 6/5 x 4/5 ) x 25/16 = ( 25/16 x 4/5 ) x 6/5 = 100/80 x 6/5 = 600/400
OT
2
21 tháng 2 2022
a: \(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\left(\dfrac{1}{8}+\dfrac{7}{8}\right)=1+1=2\)
b: \(=\left(\dfrac{1}{6}+\dfrac{4}{6}+\dfrac{7}{6}\right)+\left(\dfrac{5}{12}+\dfrac{7}{12}\right)=2+1=3\)
2 tháng 7 2021
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
NH
3
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