Tính hợp lý biểu thức \(\dfrac{12}{11}\) - \(\dfrac{-7}{19}\)+\(\dfrac{12}{19}\)
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a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)
d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
f: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=\dfrac{-5}{19}\)
i: \(=\left(\dfrac{9}{24}-\dfrac{18}{24}+\dfrac{14}{24}\right)\cdot\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{5}{24}\cdot\dfrac{6}{5}+\dfrac{1}{2}\)
=1/4+1/2=3/4
` 7/19 . 8/11 + 3/11 . 7/19 + (-12)/19 `
`= 7/19 . ( 8/11 + 3/11 ) + (-12)/19 `
`= 7/19 . 11/11 + (-12)/19`
`= 7/19 . 1 + (-12)/19 `
`= 7/19 + (-12)/19 `
`= -5/19 `
`( 3/8 + (-3)/4 + 7/12 ) : 5/6 + 1/2`
`= 3/8 + (-3)4 + 7/12 . 6/5 + 1/2`
`= ( 9+(-18) + 14)/24 . 6/5 + 1/2`
`= 5/24 . 6/5 + 1/2`
`= 1/4 + 1/2 `
`= 3/4`
g: \(=\dfrac{-3}{4}-\dfrac{1}{4}+\dfrac{5}{7}+\dfrac{2}{7}+\dfrac{3}{5}=\dfrac{3}{5}\)
h: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=-\dfrac{5}{19}\)
i: \(=\dfrac{2013}{7}\left(19+\dfrac{5}{8}-26-\dfrac{5}{8}\right)=\dfrac{2013}{7}\cdot\left(-7\right)=-2013\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
\(=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}-\dfrac{10}{3}\right)\)
\(=\dfrac{4}{19}\cdot\left(\dfrac{-10-7-40}{12}\right)=\dfrac{4}{19}\cdot\dfrac{-57}{12}=-\dfrac{3}{3}=-1\)
=\(\dfrac{4}{19}.\left(-\dfrac{5}{6}+-\dfrac{7}{12}\right)-\dfrac{40}{57}\)
=\(-\dfrac{4}{19}-\dfrac{40}{57}\)
=\(-\dfrac{52}{57}\)
\(a.\)
\(\dfrac{27}{13}-\dfrac{106}{111}+-\dfrac{5}{111}=\dfrac{27}{13}-\dfrac{106}{111}-\dfrac{5}{111}=\dfrac{27}{13}-\left(\dfrac{106+6}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
\(b.\)
\(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
\(c.\)
\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+-\dfrac{6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{25}{31}+\dfrac{6}{31}\right)=1-1=0\)
a) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}=\dfrac{27}{13}+\left(\dfrac{-106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}+-1=\dfrac{14}{13}\)
b) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\left(\dfrac{7}{19}+\dfrac{12}{19}\right)=\dfrac{12}{11}+1=\dfrac{23}{11}\)
c)\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+\dfrac{-6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{-25}{31}+\dfrac{-6}{31}\right)=1+-1=0\)
\(\dfrac{{ - 7}}{{12}} + \dfrac{5}{{12}} = \dfrac{{ - 7 + 5}}{{12}} = \dfrac{{ - 2}}{{12}} = \dfrac{{ - 1}}{{6}}\) ;
\(\dfrac{{ - 8}}{{11}} + \dfrac{{ - 19}}{{11}} = \dfrac{{ - 8 + \left( { - 19} \right)}}{{11}} = \dfrac{{ - 27}}{{11}}\)
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
\(=\dfrac{12}{11}+1=\dfrac{23}{11}\)