Question

Tìm \(n\in N\), sao cho :

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

\(b,\left(2n^2-3n+1\right)⋮\left(2n-1\right)\)

Answer 1

a.\(2n^2-3n+1=2n\times\left(n-1\right)-\left(n-1\right)=\left(2n-1\right)\times\left(n-1\right)\Rightarrow2n-1⋮n-1\)

\(\Rightarrow2\left(n-1\right)+1⋮n-1\Rightarrow1⋮n-1\Rightarrow n-1\inƯ\left(1\right)=\left\{1\right\}\Rightarrow n=2\)

b.Tách tương tự nha

Answer 2

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

vậy với mọi x thuộc N đều t/m

b) tương tự nha