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\(M=\frac{2.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}{-5.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{3.\left(\frac{1}{17}-\frac{1}{23}+\frac{1}{31}\right)}=-\frac{2}{5}+\frac{1}{3}=\frac{1}{15}.\)
`a)1 4/23 + ( 5/21-4/23)+16/21-1/2`
`=27/23+5/21-4/23+16/21-1/2`
`=(27/23-4/23)+(5/21+16/21)-1/2`
`=23/23+21/21-1/2`
`=1+1-1/2`
`=2-1/2`
`=4/2-1/2`
`=3/2`
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`b)75%-(5/2+5/3)+(-1/2)^3`
`=3/4-5/2+5/3+(-1/8)`
`=(3/4-5/2-1/8)+5/3`
`=(6/8-20/8-1/8)+5/3`
`=-15/8+5/3`
`=-45/24+40/24`
`=-5/24`
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`c)-3/4(-55/9).8/11`
`=-3/4.(-40/9)`
`=-10/3`
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`d)-3/8 . 6/13 + 7/13 . (-3/8) + 1 3/8`
`= -3/8 . (6/13 + 7/13) + 11/8`
`= -3/8 . 13/13 + 11/8`
`= -3/8 .1 + 11/8`
`= -3/8 + 11/8`
`= 8/8`
`=1`
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
\(\left\{2^3+\left[1+\left(3-1\right)^2\right]\right\}:13\)
\(=\left(8+1+2^2\right):13\)
\(=13:13\)
\(=1\)
{+ [1 + ]}: 13
= [8 + (1 + )] : 13
= [8 + (1 + 4)] : 13
= (8 + 5) : 13= 13 : 13 = 1a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2
b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2
c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)
d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45
e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)
f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) . ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)
g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)
h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)
i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\)
= \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\) = \(\dfrac{-193}{200}\)
j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )
= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\)
= -1+\(\dfrac{-13}{21}\)
= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)
Khôi nguyễn
\(=\left(8+1+4\right):13=1\)