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\(\dfrac{1}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}\)
\(=\dfrac{1}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{5}+\dfrac{1}{6}-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
\(a.\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{7+12}{19}=\dfrac{19}{19}=1\)
\(b.\dfrac{5}{13}+\dfrac{4}{13}=\dfrac{5+4}{13}=\dfrac{9}{13}\)
\(c.\dfrac{24}{11}-\dfrac{13}{11}=\dfrac{24-13}{11}=\dfrac{11}{11}=1\)
\(d.\dfrac{35}{8}-\dfrac{19}{8}=\dfrac{35-19}{8}=\dfrac{16}{8}=2\)
a,\(\dfrac{7}{19}\)+\(\dfrac{12}{19}\)=\(\dfrac{19}{19}\)=1
b,\(\dfrac{5}{13}\)+\(\dfrac{4}{13}\)=\(\dfrac{9}{13}\)
c,\(\dfrac{24}{11}\)-\(\dfrac{13}{11}\)=\(\dfrac{11}{11}\)=1
d,\(\dfrac{35}{8}\)-\(\dfrac{19}{8}\)=\(\dfrac{16}{8}\)=2
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
Tính bằng cách thuận tiện
\(\dfrac{7}{13}\) + \(\dfrac{5}{9}\) + \(\dfrac{6}{13}\) + \(\dfrac{5}{9}\)
`7/13+5/9+6/13+5/9`
`=(7/13+6/13)+(5/9+5/9)`
`=13/13+10/9`
`=1+10/9`
`=9/9+10/9`
`=19/9`
`(13+x)/20 = 3/4`
`(13+x) xx4=3xx20`
`(13+x)xx4=60`
`13+x=60:4`
`13+x=15`
`x=15-13`
`x=2`
__
`(23-x)/25 =4/5`
`(23-x)xx5=4xx25`
`(23-x)xx5=100`
`23-x=100:5`
`23-x=20`
`x=23-20`
`x=3`
a: Sửa đề; \(A=\dfrac{7.2:2\cdot28.6+1.43\cdot2\cdot64}{1+3+5+7+...+49-339}\)
\(=\dfrac{3.6\cdot28.6+2.86\cdot64}{1+3+5+...+49-339}\)
\(=\dfrac{2.86\left(64+36\right)}{25^2-339}=\dfrac{286}{286}=1\)
b: =>2(x+7/8)=6*13/4=78/4=39/2
=>x+7/8=39/4
=>x=71/8
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\(\dfrac{5}{14}\times\dfrac{7}{13}\times\dfrac{26}{25}=\dfrac{35}{182}\times\dfrac{26}{25}=\dfrac{5}{26}\times\dfrac{26}{25}=\dfrac{130}{650}=\dfrac{1}{5}\)
\(\dfrac{5}{14}\) x \(\dfrac{7}{13}\) x \(\dfrac{26}{25}\)
=\(\dfrac{5}{26}\) x \(\dfrac{26}{25}\)
=\(\dfrac{1}{5}\)
=20-13/7
=140/7-13/7
=127/7
`20/13 xx 13 - 13/7`
`= 20 - 13/7`
`= 18``1/7`