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1. ( -37 ) + 14 + 26 + 37
=(-37+37)+(14+26)
=0+30
=30
2. ( -24 )+ 6 + 10 + 24
=(-24+24)+(6+10)
=0+16
=16
3. 15 + 23 + ( -25 ) + ( -32 )
=[15+(-25)]+[23+(-32)]
=-10+(-9)
=-19
4. 60 + 33 + ( -50 ) + ( -33 )
=[60+(-50)]+[33+(-33)]
=10+0
=10
5. (- 16 ) + ( -209 ) + ( -14 ) + 209
=[-16+(-14)]+(-209+209)
=-30+0
=-30
6. (- 12 ) + ( -13 ) + 36 + ( -11 )
=[-12+(-11)]+(-13+36)
=-23+23
=0
7. - 16 + 24 - 16 - 34
=(24-34)-16-16
=-10-16-16
=-42
8. 25 + 37 - 48 - 25 - 37
=(25-25)+(37-37)-48
=0+0-48
=-48
9. 2575 + 37 - 48 - 25 - 37
=(2575-25)+(37-37)
=2550+0
=2550
10. 34 + 35 + 36 + 37 - 14 - 15 - 16 - 17
=(34-14)+(35-15)+(36-16)+(37-17)
=20+20+20+20
=20.4
=80
cái này giống BTVN của mk,nhưng dễ mak,tự làm đc mak
cố lên
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
34 +35 +36+37-14-15-16-17
=(34-14)+(35-15)+(36-16)+(37-17)
=20 +20 +20 +20
= 20*4
=80
mình nhanh nhất đi
Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
Ta có:
A=3 +32 +33 +34 + 35 +...+39
A=(3+32+33) + (34+35+36) + (37+38+39)
A= 39 + 39. 34 + 39. 37
A= 39. (1+34+37)\(⋮\)39
Vậy A\(⋮\)39
A=3 +32 +33 +34 + 35 +...+39
A=(3+32+33) + (34+35+36) + (37+38+39)
A= 39 + 39. 34 + 39. 37
A= 39. (1+34+37)\(⋮\)39
Vậy A\(⋮\)39