(23 . 1/3 - 1/3)1000 – 2.5
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a: \(\left[504-\left(25\cdot8+70\right)\right]:9-15+19^0\)
\(=\left(504-270\right):9-15+1\)
\(=29-14\)
=15
a) [504 - (25 . 8 + 70)] : 9 - 15 + 190
[504 - (25 . 8 + 70)] : 9 - 15 + 1
[504 - (200 + 70)] : 9 - 15 + 1
(504 - 270) : 9 - 15 + 1
234 : 9 - 15 + 1
= 26 - 15 + 1
= 11 + 1
= 12
b) 5 . {26 - [3 . (5 + 2 . 5) + 15] : 15}
5 . {26 - [3 . (5 + 10) + 15] : 15}
5 . [26 - (3 . 15 + 15) : 15]
5 . [26 - (45 + 15) : 15]
5 . (26 - 60 : 15)
5 . (26 - 4)
= 5 . 22
= 110
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
a) \(\dfrac{3}{8}+\dfrac{15}{-25}+\dfrac{3}{5}\)
\(=\dfrac{-9}{40}+\dfrac{3}{5}\)
\(=\dfrac{3}{8}\)
b) \(\dfrac{-5}{18}+\dfrac{23}{45}-\dfrac{9}{10}\)
\(=\dfrac{7}{30}-\dfrac{9}{10}\)
\(=\dfrac{-2}{3}\)
c) \(\dfrac{-5}{12}+\dfrac{15}{18}-2,25\)
\(=\dfrac{5}{12}-2,25\)
\(=\dfrac{-11}{6}\)
d) \(\dfrac{5}{6}+\dfrac{2}{3}-0,5\)
\(=\dfrac{3}{2}-0,5\)
\(=1\)
\(=\dfrac{5}{21}+\dfrac{16}{21}-\left(\dfrac{19}{23}+\dfrac{4}{23}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)
\(=\dfrac{5^3\cdot2^3+2\cdot5^3+5^3}{55}=\dfrac{5^3\left(2^3+2+1\right)}{55}=5^2=25\)