Question

Cho biểu thức:

\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

Tìm \(x\in Z\) để P có giá trị nguyên.

Answer 1

P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

P = \(\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

P = \(\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

P = \(\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\) = \(\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\) = \(\frac{x-4}{x-2}\) = \(\frac{x-2}{x-2}-\frac{2}{x-2}\) = \(1-\frac{2}{x-2}\)

Để P có giá trị nguyên thì x - 2 ∈ Ư(2) = {1; 2; -1; -2}

x - 2	1	2	-1	-2
x	3	4	1	0

Vậy để P nguyên thì x = {3; 4; 1; 0}