a/ \(M=1+3+3^2+.....+3^{119}\)

\(\Leftrightarrow3M=3+3^2+.....+3^{119}+3^{120}\)

\(\Leftrightarrow3M-M=\left(3+3^2+.....+3^{120}\right)-\left(1+3+....+3^{119}\right)\)

\(\Leftrightarrow2M=3^{120}-1\)

\(\Leftrightarrow M=\dfrac{3^{120}-1}{2}\)

b/ \(M=1+3+3^2+..........+3^{119}\)

\(=\left(1+3+3^2\right)+........+\left(3^{117}+3^{118}+3^{119}\right)\)

\(=1\left(1+3+3^2\right)+........+3^{117}\left(1+3+3^2\right)\)

\(=1.13+.....+3^{117}.13\)

\(=13\left(1+.....+3^{117}\right)⋮13\Leftrightarrow M⋮13\left(đpcm\right)\)

a) \(M=1+3+3^2+3^3+...+3^{119}\)

\(3M=3+3^2+3^3+3^4+...+3^{119}+3^{120}\)

\(3M-M=\left(3+3^2+3^3+...+3^{120}\right)-\left(1+3+3^2+...+3^{119}\right)\)

\(2M=3^{120}-1\)

\(M=\frac{3^{120}-1}{2}\)

b) \(M=1+3+3^2+3^3+...+3^{118}+3^{119}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{117}\right)\)chia hết cho \(13\).

\(M=1+3+3^2+3^3+...+3^{118}+3^{119}\)

\(=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)

\(=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)

\(=40\left(1+3^4+...+3^{116}\right)\)chia hết cho \(5\).

cảm ơn cô ạ

3M=3+3²+3³+3⁴+...+3¹¹⁹+3¹²⁰

3M-M=(3+3²+3³+3⁴+...+3¹¹⁹+3¹²⁰)-(1+3+3²+3³+...+3¹¹⁸+3¹¹⁹)

2M=3¹²⁰-1=>M=(3¹²⁰-1):2

a) M =  1 +3 +3² +3³ + ....+ 3¹¹⁸ +3¹¹⁹

3M= 3 +3² +3³ + ....+ 3¹¹⁹ +3¹²⁰

3M-M= (3 +3² +3³ + ....+ 3¹¹⁹ +3¹²⁰)-(1 +3 +3² +3³ + ....+ 3¹¹⁸ +3¹¹⁹)

2M= 3¹²⁰-1

M= \(\frac{3^{120}-1}{2}\)

b) M=1 +3 +3² +3³ + ....+ 3¹¹⁸ +3¹¹⁹

= (1 +3 +3² +3³ )+(3⁴+3⁵+3⁶+3⁷)+....+ (3¹¹⁷+3¹¹⁸ +3¹¹⁹)

= 40+3⁴.(1 +3 +3² +3³ )+3⁸.(1 +3 +3² +3³ )+....+3¹¹⁷.(1 +3 +3² +3³ )

= 40+3⁴.40+3⁸.40+....+3¹¹⁷.40

= 40.(1+3⁴+3⁸+....+3¹¹⁷) 

vì 40 chia hết cho 5

=> M chia hết cho 5.

M=1 +3 +3² +3³ + ....+ 3¹¹⁸ +3¹¹⁹

= (1+3+3²)+(3³+3⁴+3⁵)+....+(3¹¹⁷+3¹¹⁸+3¹¹⁹)

= 13+3³.13+3⁶+....+3¹¹⁷.13

= 13.(1+3³+3⁶+....+3¹¹⁷)

Vì 13 chia hết cho 13

=> M chia hết cho 13.

giống bạn công chúa bloom

a, \(M=2x^3+xy^2-3xy+1\)

b, Thay x = -1 ; y = 2 ta được 

M = -2 - 2 + 6 + 1 = 3 

ủng hộ mình lên 110 với các bạn

Bài 1 : \(A=1+3+3^2+...+3^{31}\)

a. \(A=\left(1+3+3^2\right)+...+3^9.\left(1.3.3^2\right)\)

\(\Rightarrow A=13+3^9.13\)

\(\Rightarrow A=13.\left(1+...+3^9\right)\)

\(\Rightarrow A⋮13\)

b. \(A=\left(1+3+3^2+3^3\right)+...+3^8.\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=40+...+3^8.40\)

\(\Rightarrow A=40.\left(1+...+3^8\right)\)

\(\Rightarrow A⋮40\)

Bài 2:

Ta có: \(C=3+3^2+3^4+...+3^{100}\)

\(\Rightarrow C=(3+3^2+3^3+3^4)+...+(3^{97}+3^{98}+3^{99}+3^{100})\)

\(\Rightarrow3.(1+3+3^2+3^3)+...+3^{97}.(1+3+3^2+3^3)\)

\(\Rightarrow3.40+...+3^{97}.40\)

Vì tất cả các số hạng của biểu thức C đều chia hết cho 40

\(\Rightarrow C⋮40\)

Vậy \(C⋮40\)

a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)

b: Thay x=11 vào M, ta được:

\(M=\dfrac{8}{11-3}=1\)

a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)

\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)

b) Thay \(x=11\left(TM\right)\) vào biểu thức M: 

\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)

Bài 4 :

Thay x=y+5 , ta có :

a ) ( y+5)*(y5+2)+y*(y-2)-2y*(y+5)+65

=(y+5)*(y+7)+y^2-2y-2y^2-10y+65

=y^2+7y+5y+35-y^2-2y-2y^2-10y+65

= 100

Bài 5 :

A = 15x-23y

B = 2x-3y

Ta có : A-B

= ( 15x -23y)-(2x-3y)

=15x-23y-2x-3y

=13x-26y

=13x*(x-2y) chia hết cho 13 

=> Nếu A chia hết cho 13 thì B chia hết cho 13 và ngược lại