3x1/2
1-(2/3+1/6)
K
Khách
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TL
3
2 tháng 8 2023
a) \(\dfrac{11}{3}-\dfrac{2}{3}\times\dfrac{5}{6}\)
\(=\dfrac{11}{3}-\dfrac{10}{18}\)
\(=\dfrac{11}{3}-\dfrac{5}{9}\)
\(=\dfrac{33}{9}-\dfrac{5}{9}\)
\(=\dfrac{28}{9}\)
b) \(\dfrac{13}{2}+\dfrac{7}{6}:\dfrac{2}{3}\)
\(=\dfrac{13}{2}+\dfrac{7}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{13}{2}+\dfrac{7}{4}\)
\(=\dfrac{26}{4}+\dfrac{7}{4}\)
\(=\dfrac{33}{4}\)
c) \(\dfrac{28}{9}-\dfrac{3}{4}+\dfrac{1}{3}\)
\(=\dfrac{28}{9}+\dfrac{3}{9}-\dfrac{3}{4}\)
\(=\dfrac{31}{9}-\dfrac{3}{4}\)
\(=\dfrac{124}{36}-\dfrac{27}{36}\)
\(=\dfrac{97}{36}\)
d) \(\dfrac{5}{8}:\dfrac{2}{3}\times\dfrac{1}{5}\)
\(=\dfrac{5}{8}\times\dfrac{3}{2}\times\dfrac{1}{5}\)
\(=\dfrac{15}{80}\)
\(=\dfrac{3}{16}\)
HQ
Hà Quang Minh
Giáo viên
2 tháng 8 2023
a) \(\dfrac{11}{3}-\dfrac{2}{3}\times\dfrac{5}{6}=\dfrac{11}{3}-\dfrac{5}{9}=\dfrac{28}{9}\)
b) \(\dfrac{13}{2}+\dfrac{7}{6}:\dfrac{2}{3}=\dfrac{13}{2}+\dfrac{7}{6}\times\dfrac{3}{2}=\dfrac{13}{2}+\dfrac{7}{4}=\dfrac{33}{4}\)
c) \(\dfrac{28}{9}-\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{97}{36}\)
d) \(\dfrac{5}{8}:\dfrac{2}{3}\times\dfrac{1}{5}=\dfrac{5}{8}\times\dfrac{3}{2}\times\dfrac{1}{5}=\dfrac{3}{16}\)
HQ
2
29 tháng 8 2019
a) 2/9 : 2/3 x 1/2
= 2/9 x 3/2 x 1/2
= 6/18 x 1/2
= 6/36
b) 2 + 1/4 x 4/3
= 2 + 4/7
= 2/1(7) + 4/7
= 14/7 + 4/7
= 18/7
c) 3 x 1/2 x 1/4
= 3/1 x 1/2 x 1/4
= 3/2 x 1/4
= 3/8
#Hok_tốt
18 tháng 8 2015
\(\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+...+\frac{1}{8}\times\frac{1}{9}=\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{8\times9}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
26 tháng 3 2019
\(\frac{1}{3}\times\frac{1}{6}\times\frac{1}{9}=\frac{1\times1\times1}{3\times6\times9}=\frac{1}{162}\)
\(\frac{1}{3}\times\frac{1}{6}:\frac{1}{9}=\frac{1}{3}\times\frac{1}{6}\times\frac{9}{1}=\frac{1\times1\times9}{3\times6\times1}=\frac{9}{18}=\frac{1}{2}\)
Tự tính câu cuối đi
LT
1
1 tháng 5 2020
\(\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{79}\cdot\frac{1}{80}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{79\cdot80}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{79}-\frac{1}{80}\)
\(=1-\frac{1}{80}\)
\(=\frac{79}{80}\)
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