Answer
`A = (1/{x - 2} - {2x}/{4 - x^2} + 1/{2 + x}) . (2/x - 1)` `(Đk: x \ne 0 ; x \ne 2 ; x \ne - 2)`
`A = (1/{x - 2} + {2x}/{x^2 - 4} + 1/{x + 2}) . (2/x - x/x)`
`A = (1/{x - 2} + {2x}/{x^2 - 2^2} + 1/{x + 2}) . ({2 - x}/x)`
`A = (1/{x - 2} + {2x}/{(x - 2) . (x + 2)} + 1/{x + 2}) . ({- (x - 2)}/x)`
`A = [{1 . (x + 2)}/{(x - 2) . (x + 2)} + {2x}/{(x - 2) . (x + 2)} + {1 . (x - 2)}/{(x - 2) . (x + 2)}] . ({- (x - 2)}/x)`
`A = [{x + 2}/{(x - 2) . (x + 2)} + {2x}/{(x - 2) . (x + 2)} + {x - 2}/{(x - 2) . (x + 2)}] . ({- (x - 2)}/x)`
`A = {x + 2 + 2x + x - 2}/{(x - 2) . (x + 2)} . ({- (x - 2)}/x)`
`A = {4x}/{(x - 2) . (x + 2)} . ({- (x - 2)}/x)`
`A = - 4/{x + 2}`
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`b,` Ta có:
`2x^2 + x = 0`
`<=> x . (2x + 1) = 0`
`<=>` $\left[\begin{matrix} x = 0\\ 2x + 1 = 0\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x = 0\\ 2x = -1\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x = 0\\ x = \dfrac{-1}{2}\end{matrix}\right.$
Thay `x = 0` vào `A` ta có:
`A = -4/{0 + 2}`
`A = -4/2`
`A = -2`
Thay `x = {-1}/2` vào `A` ta có:
$A = \dfrac{-4}{\dfrac{-1}{2} + 2}$
$A = \dfrac{-4}{\dfrac{3}{2}}$
`A = {-8}/3`
Vậy `x = 0` thì `A = -2` `,` `x = {-1}/2` thì `A = {-8}/3`
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`c,` `A = - 4/{x + 2}`
`<=> 1/2 = -4/{x + 2}`
`<=> x + 2 = -4 . 2`
`<=> x + 2 = -8`
`<=> x = -8 - 2`
`<=> x = -10`
Vậy `x = -10` thì `A = 1/2`
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`d,` `A = {-4}/{x + 2}`
Để `A` nguyên dương
`<=>` `x + 2` thuộc ước âm của `4`
Ta có:
`Ư(4) = {-1 ; -2 ; -4}`
Lập bảng giá trị:
$\begin{array}{|c|c|c|}\hline x + 2&-1&-2&-4\\\hline x&-3&-4&-6\\\hline \text{Thử lại}&\text{TM}&\text{TM}&\text{TM}\\\hline\end{array}$
Vậy `x \in {-3 ; -4 ; -6}` thì `A` nguyên dương`.`
