`A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}`
`ĐKXĐ: x+3\ne0, x^2+x-6\ne0, 2-x\ne0 ⇔ x+3\ne0, (x+3)(x-2)\ne0, 2-x\ne0 ⇔ x\ne2, x\ne-3.`
`a) A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}`
`A=\frac{(x+2)(x-2)}{(x+3)(x-2)}-\frac{5}{(x-2)(x+3)}-\frac{x+3}{(x-2)(x+3)}`
`A= \frac{(x+2)(x-2)-5-x-3}{(x+3)(x-2)}`
`A= \frac{x^2-4-5-x-3}{(x+3)(x-2)}`
`A= \frac{x^2-x-12}{(x+3)(x-2)}`
`A= \frac{x^2-4x+3x-12}{(x+3)(x-2)}`
`A= \frac{x(x-4)+3(x-4)}{(x+3)(x-2)}`
`A= \frac{(x-4)(x+3)}{(x+3)(x-2)}`
`A=\frac{x-4}{x-2}.`
Vậy `A=\frac{x-4}{x-2} (x\ne2, x\ne-3).`
`b)`Để `A>0 ⇔ \frac{x-4}{x-2} > 0.`
Ta có hai trường hợp:
`1) x-4>0, x-2>0 ⇔ x>4, x>2⇔ x>4.`
`2) x-4<0, x-2<0 ⇔ x<4, x<2⇔ x<2.`
Vậy để `A>0 ⇔ x>4` hoặc `x<2.`
`c)\frac{x-4}{x-2} = \frac{x-2-2}{x-2}=\frac{x-2}{x-2} - \frac{2}{x-2} = 1- \frac{2}{x-2}.`
Để `A>0⇔ x\ge4` hoặc `x<2.`
Có: `1∈ZZ ⇒ \frac{2}{x-2}` phải `∈ZZ.`
`⇒x-2 ∈ Ư(2)={±1;±2}.`
Ta có 4 trường hợp, ta có bảng sau:
$\left[\begin{array}{ccc}x-2&1&-1&2&-2\\x&3&1&4&0\\KL&KTM&TM&TM&TM\end{array}\right]$
Vậy để `A` nguyên dương thì `x∈{0;1;4}.`
