Question

Chứng minh rằng với mọi giá trị nguyên của n ta luôn có:

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

b) \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)

Answer 1

Ngân ơi, bài ai giao thế ?

Answer 2

a,

\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\\ =\left(n^2+3n-1\right)n+\left(n^2+3n-1\right)2-n^3+2\\ =n^3+3n^2-n+2n^2+6n-2-n^3+2\\ =5n^2+5n\\ =5\cdot\left(n^2+n\right)⋮5\\ \RightarrowĐpcm\)

b,

\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\\ =\left(6n+1\right)n+\left(6n+1\right)5-\left(3n+5\right)2n-\left(3n+5\right)\\ =6n^2+n+30n+5-6n^2-10n-3n-5\\ =18n⋮2\\ \RightarrowĐpcm\)