Đáp án + Giải thích các bước giải:
`a)` `A=(1/(x-2)-(2x)/(4-x^2)+1/(2+x))(2/x-1)(x ne pm2,x ne 0)`
`=(1/(x-2)+(2x)/(x^2-4)+1/(2+x))(2/x-x/x)`
`=(1/(x-2)+(2x)/[(x-2)(x+2)]+1/(2+x))((2-x)/x)`
`=((x+2)/[(x-2)(x+2)]+(2x)/[(x-2)(x+2)]+(x-2)/[(x-2)(x+2)])*(2-x)/x`
`=(x+2+2x+x-2)/[(x-2)(x+2)]*(2-x)/x`
`=(4x)/[(x-2)(x+2)]*(2-x)/x`
`=(-4x)/[(2-x)(x+2)]*(2-x)/x=(-4)/(x+2)`.
`b)` `2x^2+x=0`
`<=>x(2x+1)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0(ktm)\\x=-\dfrac{1}{2}(tm)\end{array} \right.\)
Thay `x=-1/2` vào biểu thức `A` ta được :
`A=(-4)/(x+2)=(-4)/[(-1/2)+2]=-8/3`
Vậy `A=-8/3`
`c)` `A=1/2<=>(-4)/(x+2)=1/2`
`<=>(-8)/[2(x+2)]=(x+2)/[2(x+2)]`
`=>-8=x+2`
`<=>-8-x-2=0`
`<=>-10-x=0<=>x=-10(tm)`
`d)` `A=(-4)/(x+2)`
`<=>x+2inƯ(-4)={pm1;pm2;pm4}`
$\begin{array}{|c||c||c|}\hline\text{x+2}&\text{1}&\text{-1}&\text{2}&\text{-2}&\text{4}&\text{-4}\\\hline\text{x}&\text{-1}&\text{-3}&\text{0}&\text{-4}&\text{2}&\text{-6}\\\hline\text{A}&\text{-4(ktm)}&\text{4(tm)}&\text{-2(ktm)}&\text{2(tm)}&\text{Không tính được}&\text{1(tm)}\\\hline\end{array}$
Vậy `x in {-3;-4;-6}` để `A` nguyên dương.
