Question

\(B=\dfrac{4}{x^2-2x+2}\)

Tìm x nguyên để B có giá trị nguyên

Answer 1

Để B là số nguyên thì \(4⋮x^2-2x+2\)

=>\(x^2-2x+2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(\left(x-1\right)^2+1\in\left\{1;-1;2;-2;4;-4\right\}\)

mà \(\left(x-1\right)^2+1>=1\forall x\)

nên \(\left(x-1\right)^2+1\in\left\{1;2;4\right\}\)

=>\(\left(x-1\right)^2\in\left\{0;1;3\right\}\)

=>\(x-1\in\left\{0;1;-1;\sqrt{3};-\sqrt{3}\right\}\)

=>\(x\in\left\{1;2;0;\sqrt{3}+1;1-\sqrt{3}\right\}\)

mà x nguyên

nên \(x\in\left\{1;2;0\right\}\)