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ủng hộ mk

tk mk

kb rồi nhé

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Code : Breacker

Ta có \(2n+1⋮n-2\)

\(\Rightarrow2.\left(n-2\right)+3⋮n-2\)

\(\Rightarrow3⋮n-2\)

\(\Rightarrow n-2\in\text{Ư}\left(3\right)\)

\(\Rightarrow n-2\in\left\{1;-1;3;-3\right\}\)

\(\Rightarrow n\in\left\{3;1;5;-1\right\}\)

\(a.b=\left(a+b\right).\left(a+b\right)\)

\(a.b=\left(a+b\right)^2\)

\(\Leftrightarrow a,b\in\left\{0\right\}\)

x+1 chia hết 2x-1

2(x+1) chia hết 2x-1

2x+2 chia hết 2x-1

2x-1+3 chia hết 2x-1

3 chia hết 2x-1

Do 2x-1 là số lẻ nên 2x-1=-3;-1;1;3

2x=-2;0;2;4

x=-1;0;1;2

b)Ghi đầu baì

=(1+2+3+...+100).(1²+2²+3²+....+100²).(65.111-13.555)

=(1+2+3+...+100).(1²+2²+3²+....+100²).(65.111-13.5.111)

=(1+2+3+...+100).(1²+2²+3²+....+100²).(111.(65-65))

=(1+2+3+...+100).(1²+2²+3²+....+100²).111.0

=(1+2+3+...+100).(1²+2²+3²+....+100²).0

=0

\(A=8+8^2+8^3+8^4+...+8^{19}+8^{20}\)

\(A=\left(8+8^2\right)+\left(8^3+8^4\right)+...+\left(8^{19}+8^{20}\right)\)

\(A=72+72\cdot8^2+...+72\cdot8^{18}\)

\(A=72\cdot\left(1+8^2+...+8^{18}\right)\)

\(A=8\cdot9\cdot\left(1+8^2+...+8^{18}\right)⋮9\left(ĐPCM\right)\)

A= 8¹+8²+8³+8⁴+...+8¹⁹+8²⁰

A= (8¹+8²)+(8³+8⁴)+...+(8¹⁹+8²⁰)

A=(8x1+8x8)+(8³x1+8³x8)+...+(8¹⁹x1+8¹⁹x8)

A=8x(1+8)+8³x(1+8)+...+8¹⁹x(1+8)

A=8x9+8³x9+...+8¹⁹x9

A=9x(8+8³+...+8¹⁹)

Vì 9 chia hết cho 9 suy ra 9x(8+8³+...+8¹⁹) chia hết cho 9 hay A chia hết cho 9

Vậy A chia hết cho 9

\(A=2+2^2+2^3+2^4+...+2^{19}+2^{20}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)

\(A=6+6\cdot2^2+...+6\cdot2^{18}\)

\(A=6\cdot\left(1+2^2+...+2^{18}\right)⋮\text{ }3\text{ v}\)

\(A=2+2^2+2^3+....+2^{20}.\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{19}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{19}.3\)

\(=3.\left(2+2^3+...+2^{19}\right)⋮3\)

\(\Rightarrow A⋮3\)( đpcm)
\(\text{Vì A có các hạng tử đều là lũy thừa của 2 nên }\) \(A⋮2\)

Vì \(A⋮2\)và \(A⋮3\)Nên \(A⋮6\)(đpcm)

hocjhocj ở mô đây