\(11.5^{2n}+3^{3n+2}+2^{3n+1}\)\(=11.25^n+8^n.4+8^n.2\)\(=11.25^n+6.8^n\)

Vì 25 = 8 (dư 17)

➩ \(11.5^{2n}+3^{3n+2}+2^{3n+1}\)\(=11.25^n+6.8^n\)\(=11.8^n+6.8^n=17.8^n=0\) (dư 17)

Hay \(11.5^{2n}+3^{3n+2}+2^{3n+1}\) ⋮ 17

a/ \(x^4+2x^3+x^2+x^2+2xy+y^2=0\)

\(\Leftrightarrow\left(x^2+x\right)^2+\left(x+y\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x=0\\x+y=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\end{matrix}\right.\)

b/ 72 chia hết 24 nên ta chỉ cần chứng minh \(A=n^3+23n⋮24\)

\(A=n^3+23n=n\left(n^2+23\right)=n\left[n^2-1+24\right]\)

\(=n\left[\left(n-1\right)\left(n+1\right)+24\right]=n\left(n-1\right)\left(n+1\right)+24n\)

\(24n\) hiển nhiên chia hết 24. Xét \(B=n\left(n-1\right)\left(n+1\right)\)

B là tích 3 số nguyên liên tiếp \(\Rightarrow B⋮3\)

n lẻ \(\Rightarrow n=2k+1\Rightarrow B=\left(2k+1\right)2k.\left(2k+2\right)\)

\(B=4k\left(k+1\right)\left(2k+1\right)\)

\(k\left(k+1\right)\) là tích 2 số nguyên liên tiếp \(\Rightarrow\) chia hết cho 2 \(\Rightarrow B⋮8\)

Mà 3;8 nguyên tố cùng nhau \(\Rightarrow B⋮24\Rightarrow A⋮24\)

Do

=> Đpcm

b)

=> đpcm

Chúc bn học tốt

8: \(10n^3-23n^2+14n-5⋮2n-3\)

\(\Leftrightarrow10n^3-15n^2-8n^2+12n+2n-3-2⋮2n-3\)

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

hay \(n\in\left\{2;1;\dfrac{5}{2};\dfrac{1}{2}\right\}\)

1)

a)2⁵¹-1

=(2³)¹⁷-1\(⋮\)2³-1=7

Vậy 2⁵¹-1\(⋮\)7

b)2⁷⁰+3⁷⁰

=(2²)³⁵+(3²)³⁵\(⋮\)2²+3²=13

Vậy 2⁷⁰+3⁷⁰\(⋮\)13

c)17¹⁹+19¹⁷

=(BS18-1)¹⁹+(BS18+1)¹⁷

=BS18-1+BS18+1

=BS18\(⋮\)18

d)36⁶³-1\(⋮\)35\(⋮\)7

Vậy 36⁶³-1\(⋮\)7

36⁶³-1

=36⁶³+1-2

=BS37-2\(⋮̸\)37

Vậy 36⁶³-1\(⋮̸\)37

e)2⁴ⁿ-1

=(2⁴)ⁿ-1\(⋮\)2⁴-1=15

Vậy 2⁴ⁿ-1\(⋮\)15