Ta có : \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)

\(=n\left(3-2n\right)-\left(3-2n\right)-n^2-5n\)

\(=3n-2n^2-3+2n-n^2-5n\)

\(=-3n^2-3\)

\(=-3\left(n^2+1\right)⋮3\)

Vậy \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)⋮3\)

Ta có \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)=3n-2n^2-3+2n-n^2-5n=-3n-3\)

mà -3n chia hết cho 3,-3 chia hết cho 3

=> biểu thức (n-1)(3-2n)-n(n+5) chia hết cho 3(đpcm)

(n-1)(3-2n)-n(n+5)

=3n-2n²-3+2n-n²-5n

=-3n²-3

vậy (n-1)(3-2n)-n(n+5)\(⋮\)3 vs mọi giá trị của n

cung hoi kho day chu

\(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)=n\left(3-2n\right)-1\left(3-2n\right)-n\left(n+5\right)\)

\(=3n-2n^2-3+2n-n^2-5n=\left(3n+2n-5n\right)-\left(2n^2+n^2\right)-3=-3n^2-3\)

\(=-\left(3n^2+n\right)=-3n\left(n+1\right)=3.\left(-n\right).\left(n+1\right)\) chia hết cho 3 với mọi n

 n(2n-3)-2n(n+1) 
=2n^2-3n-2n^2-2n 
=-5n 
-5n chia het cho 5 voi moi so nguyên n vi -5 chia het cho 5 
vay n(2n-3)-2n(n+1) chia het cho 5

Ta có: \(n\left(2n-3\right)-2n\left(n+1\right)\) = \(2n^2-3n-2n^2-2n\)

= \(-5n\)

Vì \(-5⋮5\) => -5n \(⋮\) 5

=> \(n\left(2n-3\right)-2n\left(n+1\right)\) \(⋮\) 5 với mọi n \(\in\) Z

n(2n-3)-2n(n+1)=2n²-3n+2n²-2n=-5n \(⋮\) 5 với mọi n

xem lại câu a nhé bạn

\(n\left(2n-3\right)-2n\left(n+1\right)\)

\(=2n^2-3n-2n^2-2n\)

\(=-5n\)

\(-5n\)chia hết cho \(5\)với mọi số nguyên \(n\)vì \(-5\)chia hết cho \(5\)

Vậy : \(n\left(2n-3\right)-2n\left(n+1\right)\)chia hết cho \(5\)

\(S=\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1\)

\(=2n\left(n^2-3n-1\right)+\left(n^2-3n-1\right)-2n^3+1\)

\(=2n^3-6n^2-2n+n^2-3n-1-2n^3+1\)

\(=\left(2n^3-2n^3\right)-\left(6n^2-n^2\right)-\left(2n+3n\right)-1+1\)

\(=-5n^2-5n=-5n\left(n+1\right)⋮5\)

\(S=\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1\)

\(=2n^3-6n^2-2n+n^2-3n-1-2n^3+1\)

\(=-5n^2-5n=-5n\left(n+1\right)⋮5\)

Vậy \(\left(2n+1\right)\left(n^2-3n-1\right)-2n^3+1⋮5\)

\(b.\)\(\left(2n-1\right)^3-\left(2n-1\right)=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)

\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)

\(\text{Áp dụng hằng đẳng thức }\)\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

\(=\left(2n-1\right)\left(2n-2\right).2n=\left(2n-1\right).2\left(n-1\right).2n\)

\(=\left(2n-1\right).4.n\left(n-1\right)\)

\(n\left(n-1\right)⋮2\)(vì là tích 2 số liên tiếp)

\(\Rightarrow\left(2n-1\right).4.n\left(n-1\right)⋮\left(4.2\right)=8\)

\(\left(2n-1\right).4.n\left(n-1\right)⋮8\RightarrowĐPCM\)