b) Để 4x + 19 chia hết cho x + 1 thì 15 chia hết cho x + 1

--> x + 1 là ước của 15

TH1: x + 1 = 15 <=> x = 14

TH2: x + 1 = 1 <=> x = 0

TH3: x + 1 = 3 <=> x = 2

TH4: x + 1 = 5 <=> x= 4

\(A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8\)

\(A=\left(7+7^3\right)+\left(7^2+7^4\right)+\left(7^5+7^7\right)+\left(7^6+7^8\right)\)

\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)

\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)

\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)

\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)

Ta có: 5 ⋮ 5

⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm) 

A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78

A =  (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)

A = 7.(1 + 72)  + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)

A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)

A = 7.50 + 72.50 + 75.40 + 76.50

A = 50.(7 + 72 + 75 + 76)

Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm

A = 7 + 7² + 7³ + 7⁴ + 7⁵ + 7⁶ + 7⁷ + 7⁸

A =  (7 + 7³) + (7²+ 7⁴) + (7⁵ + 7⁷) + (7⁶ + 7⁸)

A = 7.(1 + 7²)  + 7².(1 + 7²) + 7⁵.(1 + 7²) + 7⁶.(1 + 7²)

A = 7.( 1 + 49) + 7².( 1 + 49) + 7⁵.(1 + 49) + 7⁶. (1 + 49)

A = 7.50 + 7².50 + 7⁵.50 + 7⁶.50

A = 50.(7 + 7² + 7⁵ + 7⁶)

Vì 50 ⋮ 5 nên A = 50.(7 + 7² + 7⁶) ⋮ 5 đpcm

\(A=\left(1+7\right)+...+7^{2020}\left(1+7\right)=8\left(1+...+7^{2020}\right)⋮8\)

\(A = (1 + 7) +...+7^2\)\(^0\)\(^2\)\(^0\) \((1 + 7) = 8 (1+...+7^2\)\(^0\)\(^2\)\(^0\)\() \) ⋮\(8\)

Bài 1:

a. $=(-25)(-4)(-35)=100(-35)=-3500$

b. $=16-10=6$

c. $=180-(-16)-(-36)=180+16+36=232$

d. $=250-200:[1(-3)^2+(-8)]$

$=250-200:(9-8)=250-200=50$

2.

$60+2(12-x)=-48$

$2(12-x)=60-(-48)=60+48=108$

$12-x=108:2=54$

$x=12-54=-42$
 

Bài 3:

$A=(7+7^2)+(7^3+7^4)+....+(7^{17}+7^{18})$

$=7(1+7)+7^3(7+1)+...+7^{17}(1+7)$

$=(1+7)(7+7^3+....+7^{17})=8(7+7^3+....+7^{17})\vdots 8$

b.

$A=(7+7^2+7^3)+(7^4+7^5+7^6)+...+(7^{16}+7^{17}+7^{18})$

$=7(1+7+7^2)+7^4(1+7+7^2)+....+7^{16}(1+7+7^2)$

$=(1+7+7^2)(7+7^4+....+7^{16})$

$=57(7+7^4+...+7^{16})\vdots 57$

c.

$A=(7+7^2+7^3+7^4)+(7^5+7^6+7^7+7^8)+(7^9+7^{10}+7^{11}+7^{12})+(7^{13}+7^{14}+7^{15}+7^{16})+(7^{17}+7^{18})$
$=7(1+7+7^2+7^3)+7^5(1+7+7^2+7^3)+7^9(1+7+7^2+7^3(+7^{13}(1+7+7^2+7^3)+7^{17}(1+7)$

$=(1+7+7^2+7^3)(7+7^5+7^9+7^{13})+8.7^{17}$

$=400(7+7^5+7^9+7^{13})+8.7^{17}$

Ta thấy $400(7+7^5+7^9+7^{13})\vdots 50$ nhưng $8.7^{17}\not\vdots 50$ nên $A\not\vdots 50$

Bài 1:

a. $=(-25)(-4)(-35)=100(-35)=-3500$

b. $=16-10=6$

c. $180-(-16)-(-36)=180+16+36=196+36=232$

d. $=250-200:[2000.(-3).2-6]$

$=250-200:[2000.(-6)+(-6)]$

$=250-200:[(-6)(2000+1)]=250-200[(-6).2001]$

$=250+200.6.2001=250+2401200=2401450$

Bài 2:

$60+2(12-x)=-48$

$2(12-x)=-48-60=-108$
$12-x=-108:2=-54$

$x=12-(-54)=66$

Bài 3:

Số số hạng của A: $(718-71):1+1=648$

$A=(718+71)\times 648:2=789\times 324$

a. Có: $A=789\times 324=789\times 81\times 4$

$\Rightarrow A\not\vdots 8$ (bạn xem lại đề)

b. $A=789\times 81\times 4\not\vdots 57$ (bạn xem lại đề)

c. $A=789\times 81\times 4\not\vdots 50$ 

Vậy có vẻ đề câu này sai rồi. Bạn xem lại.

71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 + 79

= ( 71 + 79 ) + ( 72 + 78 ) + ( 73 + 77 ) + ( 74 + 76 ) + 75

=     150           +           150              +      150                +          150           +         75

=                                       150        x          4         +     75 

=                                                   600       +      75

=                                                         675

Tìm x :

X x 7 + X x 2 = 108 

X  x ( 7 + 2 ) =  108

X  x    9         =   108

X                   =   108 : 9

X                   =      12

  \(71+72+73+74+75+76+77+78+79\)

\(=\left(71+79\right)+\left(72+78\right)+\left(73+77\right)+\left(74+76\right)+75\)

\(=150+150+150+150+75\)

\(=600+75\)

\(=675\)

\(X\)x\(7+X\)x\(2=108\)

   \(X\)x\(\left(7+2\right)=108\)

                  \(X\)x\(9=108\)

                          \(X=108:9\)

                         \(X=12\)

                 Vậy \(X=12\)

\(71+72+73+74+75+76+77+78+79\)

\(=\left(71+79\right)+\left(72+78\right)+\left(73+77\right)+\left(74+76\right)+75\)

\(=150+150+150+150+75\)

\(=150.4+75=600+75=675\)

\(7x+2x=108\Leftrightarrow x\left(7+2\right)=108\Leftrightarrow9x=108\)

\(\Leftrightarrow x=108:9\Leftrightarrow x=12\)

a: \(B=3^1+3^2+...+3^{2010}\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{2009}\right)⋮4\)

\(B=3\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2008}\right)⋮13\)

b: \(C=5^1+5^2+...+5^{2010}\)

\(=5\left(1+5\right)+...+5^{2009}\left(1+5\right)\)

\(=6\left(5+...+5^{2009}\right)⋮6\)

\(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)\)

\(=31\left(5+...+5^{2008}\right)⋮31\)

c: \(D=7\left(1+7\right)+...+7^{2009}\left(1+7\right)\)

\(=8\left(7+...+7^{2009}\right)⋮8\)

\(D=7\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{2008}\right)⋮57\)

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

71. The Olympic Games are the international events which take place every four years in different cities.

72. Taxi is difficult to get during rush hour.

73. Can ditionaries be used by students?

74. I am used to getting up early to learn English.

75. Television progams are interesting to watch.

76. After they had carried the Olympic torch into the stadium, games began.

77. When I got to the theater, the movie had already started.

78. How long have you learnt English?

79. The girl whose parents live in Hanoi invited me to spend my summer holidays there.

I0. The man who is in black uniform is the referee. 

81. Try to use your goods as economically as you can.

82. It is one of the most expensive cars in the world.

83. The games are not exciting enough to attract many children. 

84. Mr. Nam is said to be a famous doctor.

85. He is not intelligent enough to understand what you say.