Đáp án:
`a)` `A={x-1}/{x-3}` với `x\ne 2;x\ne 3`
`b)` `x\in {1;2;4;5}`
`c)` `x<3`
`d)` `x>3` hoặc `x<1`
Giải thích các bước giải:
`a)` Ta có: `(x-2)(x-3)=x^2-3x-2x+6`
`=x^2-5x+6`
$\\$
`A={2x-9}/{x^2-5x+6}-{x+3}/{x-2}-{2x-1}/{3-x}`
`\qquad (ĐKXĐ: x\ne 2;x\ne 3)`
`={2x-9}/{(x-2)(x-3)}-{x+3}/{x-2}+{2x-1}/{x-3}`
`={2x-9-(x+3)(x-3)+(2x-1)(x-2)}/{(x-2)(x-3)}`
`={2x-9-(x^2-9)+2x^2-4x-x+2}/{(x-2)(x-3)}`
`={x^2-3x+2}/{(x-2)(x-3)}`
`={x^2-2x-x+2}/{(x-2)(x-3)}`
`={x(x-2)-( x-2)}/{(x-2)(x-3)}`
`={(x-2)(x-1)}/{(x-2)(x-3)}={x-1}/{x-3}`
Vậy `A={x-1}/{x-3}` với `x\ne 2;x\ne 3`
$\\$
`b)` `A={x-1}/{x-3}={x-3+2}/{x-3}`
`={x-3}/{x-3}+2/{x-3}=1+2/{x-3}`
Để `A\in ZZ=>1+2/{x-3}\in ZZ`
`=>2/{x-3}\in ZZ`
`=>x-3\in Ư(2)={-2;-1;1;2}`
`=>x\in {1;2;4;5}`
Vậy `x\in {1;2;4;5}` thì `A\in ZZ`
$\\$
`c)` Để `A<1`
`<=>{x-1}/{x-3}<1` `(x\ne 2;x\ne 3)`
`<=>{x-1}/{x-3}-1<0`
`<=>{x-1-(x-3)}/{x-3}<0`
`<=>2/{x-3}<0`
`<=>x-3<0` (vì `2>0)`
`<=>x<3`
Vậy `x<3` thì `A<1`
$\\$
`d)` Để `A>0`
`<=>{x-1}/{x-3}>0`
`<=>`$\left[\begin{array}{l}\begin{cases}x-1>0\\x-3>0\end{cases}\\\begin{cases}x-1<0\\x-3<0\end{cases}\end{array}\right.$`<=>`$\left[\begin{array}{l}\begin{cases}x>1\\x>3\end{cases}\\\begin{cases}x<1\\x<3\end{cases}\end{array}\right.$
`=>`$\left[\begin{array}{l}x>3\\x<1\end{array}\right.$
Vậy `x>3` hoặc `x<1` thì `A<0`
