`a) ĐKXĐ: x-2\ne0, 4-x^2\ne0, 2+x\ne0, x\ne0 ⇔ x\ne0, ±2. `
`B=(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{x+2})(2/x-1)`
`B=[\frac{x+2}{(x-2)(x+2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}](2/x-x/x)`
`B=[\frac{x+2+2x+x-2}{(x-2)(x+2)}].\frac{2-x}{x}`
`B=\frac{4x}{(x-2)(x+2)}.\frac{-(x-2)}{x}`
`B=\frac{-4}{x+2}.`
Vậy `B=\frac{-4}{x+2} ( ∀ x\ne0, ±2 ) .`
`b) 2x^2+x=0`
`⇔x(2x+1)=0`
`⇒`\(\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0 (ktm)\\x=\frac{-1}{2}(tm)\end{array} \right.\)
Với `x=-1/2` ta có:
`B=\frac{-4}{-1/2+2}`
`B=\frac{-4}{3/2}`
`B=(-4):3/2`
`B=-4 . 2/3`
`B= -8/3.`
Vậy với `x=-1/2` thì `B= -8/3.`
`c) B=1/2 ⇔ \frac{-4}{x+2} = 1/2`
`⇔ (-4).2=(x+2).1`
`⇔ -8 =x+2`
`⇔ -8 -2 =x`
`⇔ -10 =x (tm)`
Vậy để `B=1/2 ⇔ x=-10.`
`d) ` Để `B\ge0 ⇔ \frac{-4}{x+2}\ge0 `
Có: `-4<0`
`⇒ x+2 < 0`
`⇔ x < 0-2`
`⇔x<-2.`
Để B nguyên dương thì `x+2 ∈ Ư(-4)={±1;±2;±4}`
Mà để `x` nguyên dương thì `x<-2⇒ `x+2 ={ -1;-2;-4}`
`⇔ x∈ { -3 ; -4 ; -6}.`
Vậy để `B` nguyên dương thì `x∈ { -3 ; -4 ; -6}.`
