Câu 1:
\(\Leftrightarrow2n^2-4n+5n-10+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
Câu 2:
b: \(\dfrac{x^4-4x^2+2x-4a}{x-2}=\dfrac{x^4-2x^3+2x^3-4x^2+2x-4+4-4a}{x-2}\)
\(=x^3+2x^2+2+\dfrac{4-4a}{x-2}\)
Để dưlà -23 thì 4-4a=-23
=>4a=27
=>a=27/4