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â) (81+19)+ 243
= 100 + 243
= 343
b) 168+ 79+132
= (168+132)+79
= 300 +132
=432
c) 5.25.2.16.4
=(5.2).(25.4).16
= 10.100.16
= 1000.16
= 16000
d) 32.17+32.53
= 32.(47+53)
= 32.100
= 3200
Bài 3 :
A = 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33
=> A = ( 33 + 26 ) . 8 : 2 = 236
Vậy A = 236
\(\text{#Hok tốt!}\)
a) 2 . 31 . 12 + 4 . 6 . 42 + 8 . 27 . 3
= 24 . 31 + 24 . 42 + 24 . 27
= 24 . ( 31 + 42 + 27 )
= 24 . 100
= 2400
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
\(a,19^{2018}+13^{2018}\)
\(19\equiv-1\left(mod10\right)\)
\(\Rightarrow19\equiv\left(-1\right)^{2018}=1\left(mod10\right)\)
\(13^{2018}=\left(13^2\right)^{1009}=169^{1009}\)
\(169\equiv-1\left(mod10\right)\)
\(\Rightarrow169^{1009}\equiv\left(-1\right)^{1009}=-1\left(mod10\right)\)
\(\Rightarrow19^{2018}+13^{2018}\equiv1+\left(-1\right)=0\left(mod10\right)\)
\(\Leftrightarrow19^{2018}+13^{2018}⋮10\left(đpcm\right).\)
\(b,17^{2013}+23^{2017}\)
\(17^{2013}=\left(17^2\right)^{1006}.17=289^{1006}.17\)
\(289\equiv-1\left(mod10\right)\)
\(\Rightarrow289^{1006}\equiv\left(-1\right)^{1006}=1\left(mod10\right)\)
\(17\equiv7\left(mod10\right)\)
\(\Rightarrow289^{1006}.17\equiv1.7=7\left(mod10\right)\)( 1 )
\(23^{2017}=\left(23^2\right)^{1008}.23=529^{1008}.23\)
\(529\equiv-1\left(mod10\right)\)
\(\Rightarrow529^{1008}\equiv\left(-1\right)^{2018}=1\left(mod10\right)\)
\(23\equiv3\left(mod10\right)\)
\(\Rightarrow529^{1008}.23\equiv1.3=3\left(mod10\right)\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow17^{2013}+23^{2017}\equiv7+3=10\left(mod10\right)\)
Mà \(10⋮10\Rightarrow17^{2013}+23^{2017}\equiv0\left(mod10\right)\)
\(\Leftrightarrow17^{2013}+23^{2017}⋮10\left(đpcm\right).\)
\(c,17^5+24^4-13^{21}\)
\(=\overline{...7}+\overline{...6}-\overline{...3}\)
\(=\overline{...0}⋮10\)
\(\Rightarrow17^5+24^4-13^{21}⋮10\left(đpcm\right).\)
Bài 4:
a; 12 - (10 - 19)
= 12 - (-9)
= 12 + 9
= 21
b; (-27) - (13 - 19)
= 27 - 16
= 11
Bài 5:
a; 321 + (-15) + 30 + (-321)
= [321 - 321] + (30 - 15)
= 0 + 15
= 15
b; 2018 + 432 + 168 + (-2018)
= [2018 - 2018] + (432 + 168)
= 0 + 600
= 600