cho B =3+33+35+...+31991
CMR: B:13;41
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\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{1989}+3^{1990}+3^{1991}\right)\\ A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+3^4+...+3^{1989}\right)\\ A=13\left(3+3^4+...+3^{1989}\right)⋮13\)
\(C=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\\ C=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{1986}\left(3+3^3+3^5\right)\\ C=\left(3+3^3+3^5\right)\left(1+3^6+...+3^{1986}\right)\\ C=273\left(1+3^6+...+3^{1986}\right)\\ C=13\cdot21\left(1+3^6+...+3^{1986}\right)⋮13\\ C=\left(3+3^3+3^5+3^7\right)+\left(3^9+3^{11}+3^{13}+3^{15}\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\\ C=\left(3+3^3+3^5+3^7\right)+3^8\left(3+3^3+3^5+3^7\right)+...+3^{1984}\left(3+3^3+3^5+3^7\right)\\ C=\left(3+3^3+3^5+3^7\right)\left(1+3^8+...+3^{1984}\right)\\ C=2460\left(1+3^8+...+3^{1984}\right)\\ C=41\cdot60\left(1+3^8+...+3^{1984}\right)⋮41\)
a, 4x+18:2 = 13
4x+9 = 13
x = 1
b, 48 – 3(x+5) = 24
3(x+5) = 24
x + 5 = 8
x = 3
c, 2 x - 2 0 = 3 5 : 3 3
2x – 1 = 3 2
2x = 10
x = 5
d, (15+x):3 = 3 15 : 3 12
(15+x):3 = 3 3
(15+x):3 = 27
15 + x = 54
x = 39
a) 48*23-8*6*38
=48*23-48*38
=48(23-38)
=48*(-15)=-720
b)73-9*(-15+8)
=73+135-72=1+135=136
c) 29(19-13)-19(29-13)
=29*19-29*13-19*29+19*13
=13(-29+19)=13*(-10)=-130
d) 13*(23+22)-3(17+28)
=13*45-3*45
=45(13-3)=45*10=450
e) -35+35*(-26)+35*(-73)
=35*(-1)+35*(-26)+35*(-73)
=35(-1-26-73)
=35*(-100)=-3500
f) (-26)*38+45*(-26)+(-26)*17
=(-26)(38+45+17)
=(-26)*100
=-2600
g)33*(16-9)-16*(33-9)
=33*16-33*9-16*33+16*9
=9(-33+16)=9*(-17)=-153
Ta có:
\(A=3+3^2+3^3+3^4+3^5+3^6\)
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)\)
\(A=39+3^3.\left(3+3^2+3^3\right)\)
\(A=39+3^3.39\)
\(A=39.\left(1+3^3\right)\)
Vì \(39⋮13\) nên \(39.\left(1+3^3\right)⋮13\)
Vậy \(A⋮13\)
\(#WendyDang\)
Lời giải:
$A=(3+3^2+3^3)+(3^4+3^5+3^6)$
$=3(1+3+3^2)+3^4(1+3+3^2)=(1+3+3^2)(3+3^4)=13(3+3^4)\vdots 13$
Ta có đpcm.
a) ( -3 ) + ( 8 ) + ( -13 ) + 18 + ... + ( -53 ) + 58
= [ ( -3 ) + 8 ] + [ ( -13 ) + 18 ] + ... + [ ( -53 ) + 58 ] có 6 cặp
= 5 + 5 + ... + 5
= 5 x 6
= 30
b) ( -40 ) + ( -39 ) + ... + 33 + 34 + 35
= [ ( -35 ) + 35 ] + [ ( -34 ) + 34 ] + ... + [ ( -1 ) + 1 ] + 0 + ( -40 ) + ( -39 ) + ... + ( -36 )
= 0 + 0 + ... + 0 + 0 + ( -190 )
= -190