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X + 5/9 = 4/3
X = 4/3 - 5/9
X = 7/9
X - 4/9 = 1/2
X = 1/2 + 4/9
X = 17/18
6/13 + X = 7/6
X = 7/6 - 6/13
X = 55/78
13/5 - X = 5/6
X = 13/5 - 5/6
X= 53/30
X + 5/9 = 4/3
x = 4/3 - 5/9
x = 7/9
X - 4/9 = 1/2
x = 1/2 + 4/9
x = 17/18
6/13 + X = 7/6
x = 7/6 - 6/13
x - 55/78
13/5 - X = 5/6
x = 13/5 - 5/6
x = 53/30
1) 1/3 x 1/2 x 3/7 = 3/42 = 1/14
2) 5/4 x 1/3 +1/7 = 5/12 + 1/7 = 35/84 + 12/84 = 47/84
3) 8 x ( 8/9 - 2/3 ) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 240/240 = 1
5) ( 2/5 + 3/4 ) + 8 = 23/20 + 8 = 23//20 + 160/20 = 183/20
6) 10 x ( 1/2 - 1/5 ) = 10 x 3/10 = 10/1 x 3/10 = 30/10 = 3
a. 3/5 -1/5 x 10/13
= 3/5 - 2/13
= 29/65
b. 1/2 + 3/8 : 3/4
= 1/2 + 1/2
= 1
c. (3/5 - 3/20) x4/5
= 9/20 x 4/5
= 9/25
\(a,\frac{3}{5}-\frac{1}{15}x\frac{10}{13}\)
=3/5-2/39
=107/195
b,1/2+3/8:3/4
=1/2+1/2
=1
c,(3/5-3/20)x4/5
=9/20x4/5
=9/25
a, 1/2 - 1/3 b, 1/5 - 1/7 c, 7/9 - 2/5
d, 1/2 + 1/3 e, 2/5 + 1/4 + 1/3 g,17/42 + 3/7
h, 16/17 - 12/13
`1/2 -1/3= 3/6 -2/6=1/6`
`1/5 -1/7= 7/35 - 5/35=2/35`
`7/9 -2/5= 35/45 - 18/45=17/45`
`1/2 + 1/3 = 3/6 + 2/6= 5/6`
`2/5 + 1/4 + 1/3 = 24/60 + 15/60 + 20/30=59/30`
`17/42 + 3/7= 17/42 + 18/42=35/42=5/6`
`16/17 - 12/13=4/221`
1)
a) 21/23 x 5/11 x 4/35
= \(\frac{21.5.4}{23.11.35}\)
= \(\frac{7.3.5.4}{23.11.5.7}\)
= \(\frac{3.4}{23.11}\) ( ta đã rút gọc 7 và 5 )
= \(\frac{12}{253}\)
b) 26 x 13/42 : 13
= \(\frac{26.13}{42.13}\)
= \(\frac{13.2.13}{21.2.13}\)
= \(\frac{13}{21}\)
c) 5/9 : 2/9 + 4/9 : 2/3
= \(\frac{5.9}{9.2}+\frac{4.3}{9.2}\)
= \(\frac{5}{2}+\frac{2.2.3}{3.3.2}\)
= \(\frac{5}{2}+\frac{2}{3}\)
= \(\frac{19}{6}\)
2 ) tìm x
a ) 2 - x = 1/3 + 1/6
2 - x = 1/2
x = 2 - 1/2
x = 3/2
b ) x + 5/8 = 1 + 3/4
x + 5/8 = 7/4
x = 7/4 - 5/8
x = 9/8
3)
a) x < 3/4 + 4/5
=> x < 31/20
=> x = 1
b) 1/2 + 3/5 < x < 5/4
=> 11/10 < x < 5/4
ta quy đồng phân số 11/10 và 5/4
\(\frac{11}{10}=\frac{11.4}{10.4}=\frac{44}{40}\)
\(\frac{5}{4}=\frac{5.10}{4.10}=\frac{50}{40}\)
=> 44/40 < x < 50/40
=> x không thỏa mãn
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, \(\frac{2}{5}+\frac{6}{9}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{2}{5}+\frac{2}{3}+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\frac{1}{5}\left(2+3\right)+\frac{1}{3}\left(2+1\right)+\frac{1}{4}\left(3+1\right)\)
\(=\frac{1}{5}\cdot5+\frac{1}{3}\cdot3+\frac{1}{4}\cdot4\)
\(=1+1+1\)
\(=3\)