Question

Tìm giá trị x nguyên để M, N nguyên:

a, M= \(\dfrac{3x^2-x+3}{3x+2}\)

b, N= \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\)

Answer 1

a: Để M là số nguyên thì \(3x^2+2x-3x-2+5⋮3x+2\)

\(\Leftrightarrow3x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;1\right\}\)

b: \(N=\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2+4\right)}{\left(x-2\right)^2\cdot\left(x^2+4\right)}=\dfrac{x+2}{x-2}\)

Để N là số nguyên thì \(x-2+4⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)