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a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
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\(1\frac{4}{25}+\frac{7}{21}-\frac{4}{25}+0,5+\frac{14}{21}\)
\(=\left(1\frac{4}{25}-\frac{4}{25}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)+0,5\)
\(=1+0+0,5\)
\(=1,5\)
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a)
1/12 + 1/6 + 1/2 = (1+2+6)/12 = 9/12 = 3/4
1/30 + 1/20 = (2+3)/60 = 5/60 = 1/12
1/56 + 1/42 = 1/7(1/8+1/6) = 1/7 .(3+4)/24 = 1/24
8/9- 1/72 = (8.8 - 1)/72 = 63/72 = 7/8
1/12 + 1/24 = (2+1)/24 = 1/8
7/8 - 1/8 = 6/8 = 3/4
3/4 - 3/4 = 0
b)
\(0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}\)
\(=\frac{7}{5}+\frac{3}{5}\)
\(=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.A=-1/2-1/6-1/12-1/20-1/30-1/42-1/56-1/72=-151/180
Vậy A=151/180
b.B=0,5+0,4+1/3+1/6+5/7-4/35=2
Vậy B=2
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a) 0,4(3) = \(\frac{4,\left(3\right)}{10}=\frac{4+\frac{1}{3}}{10}=\frac{13}{30}\); 0,6(2) = \(\frac{6,\left(2\right)}{10}=\frac{6+\frac{2}{9}}{10}=\frac{56}{90}=\frac{28}{45}\); 0,5(8) = \(\frac{5,\left(8\right)}{10}=\frac{5+\frac{8}{9}}{10}=\frac{53}{90}\)
Vậy A = \(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{53}{90}}:\frac{2700}{53}\) = \(\frac{13}{30}+\frac{14}{9}-\frac{5}{6}.\frac{90}{53}.\frac{53}{2700}=\frac{13}{30}+\frac{14}{9}-\frac{1}{36}=\frac{353}{180}\)
b) 0,(5) = 5/9; 0,(2) = 2/9
B = \(\left(\frac{5}{9}.\frac{2}{9}\right):\left(\frac{10}{3}.\frac{25}{33}\right)-\left(\frac{2}{5}.\frac{4}{3}\right):\frac{4}{3}\)
B = \(\frac{10}{81}.\frac{3.33}{10.25}-\frac{2}{5}=\frac{11}{225}-\frac{2}{5}=-\frac{79}{225}\)
tính hợp lý lớp 7: [(-25) . 0,27 . 0,4 ] - ( 0,5 . 0,73 . 20 ]
Giải:Ta có:
[(-25) . 0,27 . 0,4] - (0,5 . 0,73 . 20)
=[(-10) . 0,27] - (10 . 0,73)
=(-2,7) - 7,3
=-10