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Bài 1
1/2 x 3/4 : 4/5
= 3/8 : 4/5
= 3/8 x 5/4
= 15/32
2/3 + 1/6 - 1/2
= 4/6 + 1/6 - 3/6
= 5/6 - 3/6
= 1/3
3/5 + 4 : 2/3 - 3/2
= 3/5 + 4 x 3/2 - 3/2
= 3/5 + 6 - 3/2
= 3/5 + 30/5 - 3/2
= 33/5 - 3/2
= 66/10 - 15/10
= 28/5
Bài 2
x - 1/4 = 1/2
x = 1/2 + 1/4
x = 2/4 + 1/4
x = 3/4
2/3 + x = 1
x = 1 - 2/3
x = 3/3 - 2/3
x = 1/3
4 x 2 x 5 = ?
Cách 1: 4 x 2 x 5 = (4 x 2) x 5 = 8 x 5 = 40
Cách 2: 4 x 2 x 5 = 4 x (2 x 5) = 4 x 10 = 40
7 x 2 x 3 = ?
Cách 1: 7 x 2 x 3 = (7 x 2) x 3 = 14 x 3 = 42
Cách 2: 7 x 2 x 3 = 7 x (2 x 3) = 7 x 6 = 42
6 x 3 x 3 = ?
Cách 1: 6 x 3 x 3 = (6 x 3) x 3 = 18 x 3 = 54
Cách 2: 6 x 3 x 3 = 6 x (3 x 3) = 6 x 9 = 54
6 x 2 x 4 = ?
Cách 1: 6 x 2 x 4 = (6 x 2) x 4 = 12 x 4 = 48
Cách 2: 6 x 2 x 4 = 6 x (2 x 4) = 6 x 8 = 48
1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14
2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84
3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1
5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5
6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3
Cách 1:
Xét số bị trừ, ta có:
(2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016 + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) + 2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Xét số trừ, ta có:
(1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016 + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016) x (2/3 + 3/4 + 4/5 + ... + 2015/2016) + 2016/2017 x (2/3 +3/4 + 4/5 + ... + 2015/2016) =
Ta thấy số bị trừ và số trừ có số hạng giống nhau là:
(2/3 +3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Nên phép trừ trên có thể viết lại:
2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016) - 2016/2017 x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= 2016/2017 x [(1/2 + 2/3 + 3/4 + ... + 2015/2016) - (2/3 +3/4 + 4/5 + ... + 2015/2016)]
= 2016/2017 x 1/2
= 1008/2017
Cách 2:
Cách 1:
Xét số bị trừ, ta có:
(2/3 + 3/4 + 4/5 + ... + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016 + 2016/2017) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
= (2/3 + 3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016) + 2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Xét số trừ, ta có:
(1/2 + 2/3 + 3/4 + ... + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016 + 2016/2017) x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= (1/2 + 2/3 + 3/4 + ... + 2015/2016) x (2/3 + 3/4 + 4/5 + ... + 2015/2016) + 2016/2017 x (2/3 +3/4 + 4/5 + ... + 2015/2016) =
Ta thấy số bị trừ và số trừ có số hạng giống nhau là:
(2/3 +3/4 + 4/5 + ... + 2015/2016) x (1/2 + 2/3 + 3/4 + ... + 2015/2016)
Nên phép trừ trên có thể viết lại:
2016/2017 x (1/2 + 2/3 + 3/4 + ... + 2015/2016) - 2016/2017 x (2/3 + 3/4 + 4/5 + ... + 2015/2016)
= 2016/2017 x [(1/2 + 2/3 + 3/4 + ... + 2015/2016) - (2/3 +3/4 + 4/5 + ... + 2015/2016)]
= 2016/2017 x 1/2
= 1008/2017
Cách 2:
Bài giải
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
a, \(\frac{4}{5}-\frac{2}{3}+\frac{1}{5}-\frac{1}{3}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)=1-1=0\)
b, \(\frac{2}{5}\text{ x }\frac{7}{4}-\frac{2}{5}\text{ x }\frac{3}{7}\)
\(=\frac{2}{5}\text{ x }\left(\frac{7}{4}-\frac{3}{7}\right)=\frac{2}{5}\text{ x }\frac{37}{28}=\frac{37}{70}\)
c, \(\frac{13}{4}\text{ x }\frac{2}{3}\text{ x }\frac{4}{13}\text{ x }\frac{3}{12}=\frac{13\text{ x }2\text{ x }4\text{ x }3}{4\text{ x }3\text{ x }13\text{ x }12}=\frac{1}{6}\)
d, \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\frac{3}{4}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{18}{21}+\frac{3}{21}\right)+\left(\frac{19}{32}+\frac{13}{32}\right)\)
\(=1+1+1\)
\(=3\)
e, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)
3/4 x 2/2 + 1/4
= 3/4 + 1/4
= 4/4 = 1
\(\dfrac{3}{4}\times\dfrac{2}{2}+\dfrac{1}{4}\)
\(=\dfrac{3}{4}\times1+\dfrac{1}{4}\\ =\dfrac{3}{4}+\dfrac{1}{4}\\ =\dfrac{4}{4}=1\)