38 - 2 ? 31 + 3
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VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
13 tháng 2 2020
a) \(47.69-31\left(-47\right)=47-69-\left(-31\right).47\)
\(=47.69+31.47=47.\left(69+31\right)=47.100=4700\)
b) \(16\left(38-2\right)-38\left(16-1\right)=16.38-16.2-38.16-\left(-38\right).1\)
\(=\left(16.38-38.16\right)-32+38=0-32+38=6\)
c) \(\left(-41\right)\left(59-2\right)+59\left(41-2\right)\) \(=\left(-41\right).59-\left[\left(-41\right).2\right]+59.41-59.2\)
\(=\left[\left(-41\right).59+59.41\right]-\left[\left(-49\right).2-51.2\right]=0-\left(-100\right).2=200\)
d) \(125.\left(-8\right)\left(-25\right)9.4.1000\div3=125.200.9.4.1000\div3\)
\(=\left(125.4\right)\left(9\div3\right)\left(200.1000\right)=500.3.200000=300000000\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
a) \(-524\cdot\left[23+\left(-45\right)\right]+524\cdot\left(-45+123\right)\)
\(=-524\cdot\left[\left(23-45\right)-\left(-45+123\right)\right]\)
\(=-524\cdot\left(23-45+45-123\right)\)
\(=-524\cdot-100\)
\(=52400\)
b) \(47\cdot69-31\cdot\left(-47\right)\)
\(=-47\cdot\left(-69-31\right)\)
\(=-47\cdot-100\)
\(=4700\)
c) \(16\left(38-2\right)-38\left(16-1\right)\)
\(=16\cdot38-16\cdot2-38\cdot16+38\)
\(=-16\cdot2+38\)
\(=-36+38\)
\(=2\)
d) \(\left(-41\right)\cdot\left(59+2\right)+59\cdot\left(41-2\right)\)
\(=-41\cdot59-41\cdot2+41\cdot59-2\cdot59\)
\(=-41\cdot2-2\cdot59\)
\(=-2\left(41+59\right)\)
\(=-2\cdot100\)
\(=-200\)
e) \(25\cdot\left(-8\right)\cdot\left(-25\right)\cdot9\cdot4\cdot100^0:3\)
\(=\left(-1000\right)\cdot\left(-25\right)\cdot9\cdot4\cdot1:3\)
\(=25000\cdot9\cdot4:3\)
\(=25000\cdot3\cdot4\)
\(=100000\cdot3\)
\(=300000\)
29 tháng 7 2018
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
NT
0
DT
0
TT
2
AH
Akai Haruma
Giáo viên
23 tháng 10
1/
$M=\frac{1}{3.8}+\frac{1}{8.13}+\frac{1}{13.18}+....+\frac{1}{33.38}$
$5M=\frac{8-3}{3.8}+\frac{13-8}{8.13}+\frac{18-13}{13.18}+...+\frac{38-33}{33.38}$
$=\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+....+\frac{1}{33}-\frac{1}{38}$
$=\frac{1}{3}-\frac{1}{38}=\frac{35}{114}$
$\Rightarrow M=\frac{35}{114}:5=\frac{7}{114}$
AH
Akai Haruma
Giáo viên
23 tháng 10
2/
$N=\frac{1}{3.10}+\frac{1}{10.17}+\frac{1}{17.24}+\frac{1}{24.31}+\frac{1}{31.38}$
$7N=\frac{10-3}{3.10}+\frac{17-10}{10.17}+\frac{24-17}{17.24}+\frac{31-24}{24.31}+\frac{38-31}{31.38}$
$=\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+\frac{1}{17}-\frac{1}{24}+\frac{1}{24}-\frac{1}{31}+\frac{1}{31}-\frac{1}{38}$
$=\frac{1}{3}-\frac{1}{38}=\frac{35}{114}$
$\Rightarrow N=\frac{35}{114}:7=\frac{5}{114}$
QB
Bài 3: Tính (Tính hợp lí nếu có thể):
a) – (+19) – (-38) + 19 – 58
b) ( 31 + 128 ) – ( 31 – 172 – 54 )
3
S
29 tháng 11 2021
a) – (+19) – (-38) + 19 – 58
= 19 - 19 + 38 - 58
= -20
b) ( 31 + 128 ) – ( 31 – 172 – 54 )
= 31 - 31 + 172 + 128 + 54
= 300 + 54 = 354
DV
0
2 tháng 11 2017
A = 1002 + 2002 + 3002 + ... + 10002
= (1 . 100)2 + (2 . 100)2 + (3 . 100)2 + ... + (10 . 100)2
= 12 . 1002 + 22 . 1002 + 32 . 1002 + ... + 102 . 1002
= 10000(12 + 22 + 32 + ... 102)
Mà theo đề bài, 12 + 22 + 32 + ... 102 = 385 nên 10000(12 + 22 + 32 + ... 102) = 10000 . 385
= 3850000
38-2=36
31+3=34
mtr