tính bằng hai cách
a) [1/4+1/7]*1/3
b) [1/4+1/7]*1/3
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a,C1
(5/6 + 5/8) x 2/3
= 35/24 x 2/3
=35/36
C2
(5/6 + 5/8) x 2/3
= 5/6x2/3 + 5/8x2/3
= 10/18 + 10/24
= 35/36
b,C1
7/5 x 3/4 - 1/2 x 3/4
= 21/20 - 3/8
= 27/40
C2
7/5 x 3/4 - 1/2 x 3/4
=3/4 x (7/5-1/2)
=3/4x9/10
=27/40
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
a) C1
\(\dfrac{1}{8}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{2}{3}=\dfrac{5}{6}\times\left(\dfrac{1}{8}+\dfrac{2}{3}\right)=\dfrac{5}{6}\times\dfrac{19}{24}=\dfrac{95}{144}\)
C2
\(\dfrac{1}{8}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{2}{3}=\dfrac{5}{48}+\dfrac{10}{18}=\dfrac{95}{144}\)
b) C1
\(\dfrac{7}{5}\times\dfrac{3}{4}-\dfrac{1}{2}\times\dfrac{3}{4}=\dfrac{3}{4}\times\left(\dfrac{7}{5}-\dfrac{1}{2}\right)=\dfrac{3}{4}\times\dfrac{9}{10}=\dfrac{27}{40}\)
C2
\(\dfrac{7}{5}\times\dfrac{3}{4}-\dfrac{1}{2}\times\dfrac{3}{4}=\dfrac{21}{20}-\dfrac{3}{8}=\dfrac{27}{40}\)
c) C1
\(\left(\dfrac{5}{6}+\dfrac{5}{8}\right)\times\dfrac{2}{3}=\dfrac{35}{24}\times\dfrac{2}{3}=\dfrac{35}{36}\)
C2
\(\left(\dfrac{5}{6}+\dfrac{5}{8}\right)\times\dfrac{2}{3}=\left(\dfrac{5}{6}\times\dfrac{2}{3}\right)+\left(\dfrac{5}{8}\times\dfrac{2}{3}\right)=\dfrac{5}{9}+\dfrac{5}{12}=\dfrac{35}{36}\)
a: =5/6(1/8+2/3)=5/6*19/24=95/144
b: =3/4(7/5-1/2)=27/40
c: =35/24*2/3=35/36
a: =5/48+10/18=95/144
b: =3/4(7/5-1/2)=3/4x9/10=27/40
c: =2/3x35/24=70/72=35/36
a:\(=\left(\dfrac{3}{4}+\dfrac{2}{4}\right):\dfrac{3}{5}=\dfrac{5}{4}\cdot\dfrac{5}{3}=\dfrac{25}{12}\)
b: \(=\left(\dfrac{5}{6}-\dfrac{3}{6}\right):\dfrac{3}{2}=\dfrac{1}{3}\cdot\dfrac{2}{3}=\dfrac{2}{9}\)
`( 2/3 xx 4/5) xx 5/6`
`= 8/15 xx 5/6`
`= 40/90`
`=4/9`
__
`( 2/3 xx 4/5) xx 5/6`
`= 2/3 xx (4/5 xx 5/6)`
`= 2/3 xx 2/3`
`= 4/9`
-------
`(1/2 + 1/3) xx 1/5`
`= ( 3/6 +2/6) xx 1/5`
`= 5/6 xx1/5`
`= 5/30`
`=1/6`
__
`(1/2 + 1/3) xx 1/5`
`=1/2 xx 1/5 + 1/3 xx 1/5`
`= 1/10 + 1/15`
`=1/6`
a)
c1
`(2/3xx4/5)xx5/6`
`=8/15xx5/6`
`=4/9`
c2
`(2/3xx4/5)xx5/6`
`=2/3xx4/5xx5/6`
`=2/3xx5/6xx4/5`
`=5/9xx4/5`
`=4/9`
b)
c1
`(1/2+1/3)xx1/5`
`=(3/6+2/6)xx1/5`
`=5/6xx1/5`
`=1/6`
c2
`(1/2+1/3)xx1/5`
`=1/2xx1/5+1/3xx1/5`
`=1/10+1/15`
`=3/30+2/30`
`=5/30=1/6`
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
a: \(=\dfrac{10-6}{15}\cdot\dfrac{1}{3}=\dfrac{4}{45}\)
b: \(=\dfrac{10+6}{15}\cdot\dfrac{1}{3}=\dfrac{16}{45}\)
c: \(=\dfrac{4+3}{6}\cdot\dfrac{4}{3}=\dfrac{7\cdot4}{6\cdot3}=\dfrac{28}{18}=\dfrac{14}{9}\)
d: \(=\dfrac{20-14}{35}\cdot\dfrac{9}{4}=\dfrac{6}{35}\cdot\dfrac{9}{4}=\dfrac{54}{140}=\dfrac{27}{70}\)
a (2/3 - 2/5 ) nhân 1/3
C1)
=4/15 x 1/3
= 4/45
C2)
2/3 x 1/3 - 2/5 x 1/3
= 2/9 - 2/15
= 4/45
b (2/3 + 2/5 )nhân 1/3
GIỐNG CÂU A
c (2/3 + 1/2 ) chia 3/4
C1) =7/6 : 3/4
= 14/9
C2)
2/3 : 3/4 + 1/2 : 3/4
= 8/9 + 2/3
= 14/9
d(4/7 - 2/5 ) chia 4/9
C1)
= 6/35 : 4/9
= 27/70
C2)
4/7 : 4/9 - 2/5 : 4/9
= 9/7 -9/10
= 27/70
Cách 1:
\(\left(\frac{1}{4}+\frac{1}{7}\right)\times\frac{1}{3}\)
\(=\frac{11}{28}\times\frac{1}{3}\)
\(=\frac{11}{84}\)
Cách 2:
\(\left(\frac{1}{4}+\frac{1}{7}\right)\times\frac{1}{3}\)
\(=\frac{1}{4}\times\frac{1}{3}+\frac{1}{7}\times\frac{1}{3}\)
\(=\frac{1}{12}+\frac{1}{21}\)
\(=\frac{11}{84}\)
a, Cách 1:
(1/4 + 1/7) x 1/3 = 11/28 x 1/3
= 11/84
Cách 2:
(1/4 + 1/7) x 1/3
= 1/4 x 1/3 + 1/7 x 1/3
= 11/84
b, làm tương tự như trên