(2)3 - 45 : (-32) + (-2019)0 . (-1)2019
=?
K
Khách
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A
1
4 tháng 12 2023
a: \(\left(-2\right)^3-45:\left(-3\right)^2+\left(-2019\right)^0\cdot1^{2019}\)
\(=-8-45:9+1\)
\(=-8-5+1\)
=-13+1
=-12
b: \(11^{25}:11^{13}-3^5:\left(1^{10}+2^3\right)-60\)
\(=11^{25-13}-3^5:3^2-60\)
\(=11^{12}-27-60\)
\(=11^{12}-87\)
LK
1
29 tháng 12 2021
\(=-8+\dfrac{5}{3}-1=-9+\dfrac{5}{3}=\dfrac{-22}{3}\)
20 tháng 2 2020
a) Ta có: \(14-\left(5-8\right)^3+\left(-3\right)\cdot5\)
\(=14+27-15=26\)
b) Ta có: \(-7\cdot15+7\cdot\left(-35\right)+\left(-1\right)^{2019}\)
\(=-7\left(15+35\right)-1=-350-1=-351\)
c) Ta có: \(-18\cdot32+\left(-18\right)\cdot45+77\cdot\left(-32\right)-77\cdot50\)
\(=-18\left(32+45\right)+77\left(-32-50\right)\)
\(=-18\cdot77+77\cdot\left(-82\right)=77\left(-18-82\right)=77\cdot\left(-100\right)=-7700\)
d) Ta có: \(104-4\cdot\left[-5\cdot8+\left(7-10\right)^3\right]\)
\(=104-4\left[-40-27\right]=104-4\cdot\left(-67\right)=104+268=372\)
HT
3
25 tháng 11 2018
câu 1 :
10 . 72 - 10 . 5 + 10 + 32
= 10( 72 - 5 + 1 ) + 32
= 10 . 45 + 32
= 450 + 482
25 tháng 11 2018
câu 2 :
54 : [ 452 - ( 2020 - 20180 . 12019 ) ]
= 54 : [ 2025 - ( 2020 - 1 . 1 ) ]
= 54 : [ 2025 - 2019 ]
= 54 : 6
= 9
CM
5 tháng 9 2018
1) 2*17*9+18*540+29*18
= 18*17+18*540+29*18
= 18*(17+540+29)
= 18*586
= 10548
2) 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
3) (2018*2019+2019*2020)*(45*120-15*360)*(1+5+9+13+17+...+2015+2019)
= (2018*2019+2019*2020)*(15*3*120-15*120*3)*(1+5+9+13+17+...+2015+2019)
= (2018*2019+2019*2020)*0*(1+5+9+13+17+...+2015+2019)
= 0
KN
14 tháng 7 2019
\(B=2019-\frac{2019}{3}-\frac{2019}{6}-\frac{2019}{10}-...-\frac{2019}{45}\)
\(\Leftrightarrow B=2019\left(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-...-\frac{1}{45}\right)\)
\(\Leftrightarrow B=2019\left[1-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\right]\)
\(\Leftrightarrow B=2019\left[1-\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{9.10}\right)\right]\)
\(\Leftrightarrow B=2019\left[1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\right]\)
\(\Leftrightarrow B=2019\left[1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\right]\)
\(\Leftrightarrow B=2019\left[1-2\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(\Leftrightarrow B=2019\left[1-2.\frac{4}{10}\right]\)
\(\Leftrightarrow B=2019\left[1-\frac{4}{5}\right]\)
\(\Leftrightarrow B=2019.\frac{1}{5}\)
\(\Leftrightarrow B=\frac{2019}{5}\)
NT
0
\(2^3-45:\left(-3\right)^2+\left(-2019\right)^0\cdot\left(-1\right)^{2019}\)
\(=8-45:9+1\cdot\left(-1\right)\)
\(=8-5+\left(-1\right)\)
\(=3+\left(-1\right)\)
\(=2\)
\(PeaGea\)
(2)3 - 45 : (-32) + (-2019)0 . (-1)2019
= 8 - 45 : 9 + 1 . (-1)
= 8 - 5 + (-1)
= 3 + (-1)
= 2