calibudaicho

\(C=1+3+3^2+3^3+...+3^{11}\\ a,C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\\ =13+3^3.\left(1+3+3^2\right)+3^6.\left(1+3+3^2\right)+3^9.\left(1+3+3^2\right)\\ =13+3^3.13+3^6.13+3^9.13\\ =13.\left(1+3^3+3^6+3^9\right)⋮13\)

Ý a phải chia hết cho 13 chứ em?

b: C=(1+3+3^2+3^3)+...+3^8(1+3+3^2+3^3)

=40(1+...+3^8) chia hết cho 40

a: C ko chia hết cho 15 nha bạn

\(b,C=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\\ =40+3^4.\left(1+3+3^2+3^3\right)+3^8.\left(1+3+3^2+3^3\right)\\ =40.\left(1+3^4+3^8\right)⋮40\)

s=2+2^2+2^3+.....+2^100

s=2.(1+2+2^2+2^3)+......+2^97.(1+2+2^2+2^3)

s=2.15+....+2^97.15

s=15.(2+....+2^97)

=> s chia het cho 15

a=3+3^2+3^3+....+3^20

a=3.(1+3)+......+3^19.(1+3)

a=3.4+.....+3^19.4

a=4.(3+.....+3^19)

vay a chia het cho 4

tìm số tự nhiên n để a= 2n+6/n+1 là số nguyên tố

C/M C\(⋮\)4

\(C=1+3+3^2+...+3^{99}⋮4\)

\(C=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)⋮4\)

\(C=\left(1+3\right)+3^2.\left(1+3\right)+...+3^{98}.\left(1+3\right)⋮4\)

\(C=4+3^2.4+...+3^{98}.4⋮4\)

\(C=4.\left(1+3^2+...+3^{98}\right)⋮4\)

C/M C\(⋮\)40

\(C=1+3+3^2+...+3^{99}⋮40\)

\(C=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)⋮40\)

\(C=\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)⋮40\)

\(C=40.1+...+3^{96}.40⋮40\)

\(C=40.\left(1+...+3^{96}\right)⋮40\)

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

giup minh voi sap phai nop roi

câu a Achia hết cho 128

b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)

=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)

=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)

=3+3^2.13+3^5.13+.........+3^58.13

=3.13.(3^2+3^5+....+3^58)

vi tich tren co thua so 13 nen tich do chia het cho 13

=

bai1

a) A=(3¹+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)

=3^1.(1+3)+...+3^59.(1+3)

=3^1.4+....+3^59.4

=4.(3^1+...+3^59)

vi tich tren co thua so 4 nen tich do chia het cho 4

Bài 2:(12a + 36b) = (12a + 12 x 3 x b) = 12( a + 3b)chia hết cho 12