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Giải:
a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\)
\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\)
\(=\dfrac{136}{9}-\dfrac{11}{2}\)
\(=\dfrac{173}{18}\)
b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\)
\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\)
\(=\dfrac{13}{9}.\dfrac{3}{4}\)
\(=\dfrac{13}{12}\)
c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\)
\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=1+1\)
\(=2\)
d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\)
\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\)
\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\)
\(=\dfrac{53}{10}\)
e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\)
\(=\dfrac{7}{5}+\dfrac{-2}{5}\)
\(=1\)
1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
a) 8 + 2. ( − 3 ) = 8 + ( - 6 ) = 8 - 6 = 2
b) 30.55 + 30.46 – 2.15
= 30 . 55 + 30 . 46 - 30
= 30 . ( 55 + 46 - 1 )
= 30 . 100 = 3000
c) 4.15 − [ 144 − ( 12 − 4 )2 ]
= 4 . 15 - [ 144 - 82 ]
= 4 .15 - [ 144 - 64 ]
= 60 - 80
= - 20
= -5 / 7 x ( 2 / 11 + 9 / 11) + 12/7
= -5/7 x 1 + 12/7
= -5/7 + 12/7
= 1
b, = ( 9/4 x 105 - 9/4 x 101) : 3/2 - 64 x ( 1/4 - 3/4)
= [ 9/4 x ( 105 - 101)] : 3/2 -64 x -1/2
= 9/4 x 4 : 3 /2 - (-32)
= 9 : 3/2 + 32
= 18/3 +32
=
Bài 1:
a, 58.32 + 58.68 - 800
= 58.(32 + 68) - 800
= 58.100 - 800
= 5800 - 800
= 5000
b, 12020 + 280 : [55 - (7 - 4)3]
= 12020 + 280 : [ 55 - 33 ]
= 12020 + 280 : [ 28]
= 12020 + 10
= 12030
c, (96 - 19 - 45) - (55 + 96 - 119)
= 96 - 19 - 45 - 55 - 96 + 119
= (96 - 96) + (119 - 19) - (45 + 55)
= 0 + 100 - 100
= 0
\(=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}\)
\(=\frac{1}{2}+0+\frac{4}{5}\)
\(=\frac{5}{10}+\frac{8}{10}=\frac{13}{10}=1\frac{3}{10}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}\)
\(=\frac{1}{2}+0+\frac{4}{5}\)
\(=\frac{13}{10}\)