`B=(\frac{x^3}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{2+x}): (x+2+\frac{10-x^2}{x-2})`
`a) ĐKXĐ: x^3-4x \ne0, 6-3x\ne0, 2+x\ne0, x-2\ne0⇔x\ne±2.`
` B=(\frac{x^3}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{2+x}):(x+2+\frac{10-x^2}{x-2})`
`B=(\frac{x.x^2}{x(x^2-4)}+\frac{3.2}{3.(2-x)}+\frac{1}{2+x}):(\frac{x^2-4}{x-2}+\frac{10-x^2}{x-2})`
`B=(\frac{x^2}{(x-2)(x+2)}-\frac{2(x+2)}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}):(\frac{x^2-4+10-x^2}{x-2})`
`B=(\frac{x^2-2x-4+x-2}{(x-2)(x+2)}):(\frac{6}{x-2})`
`B=\frac{x^2−x−6}{(x-2)(x+2)}.\frac{x-2}{6}`
`B=\frac{x^2−x−6}{6(x+2)}`
`B=\frac{x^2+2x-3x−6}{6(x+2)}`
`B=\frac{x(x+2)-3(x+2)}{6(x+2)}`
`B=\frac{(x+2)(x-3)}{6(x+2)}`
`B=\frac{x-3}{6}.`
Vậy `B=\frac{x-3}{6}.`
`b) x^2-5x+6=0`
`⇔x^2-2x-3x+6=0`
`⇔x(x-2)-3(x-2)=0`
`⇔(x-2)(x-3)=0`
`⇒` \(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Ta thấy `x=2` $KTMĐKXĐ.$
Với `x=3` ta có `B=\frac{3-3}{6}=B=\frac{0}{6}=0.`
`c) Để B∈ZZ ⇔\frac{x-3}{6}∈ZZ`
Đặt `\frac{x-3}{6}=k (∈ZZ) `
`⇔x-3=6k`
`⇔x=6k+3`
Vậy `x` có dạng `6k+3` thì `B∈ZZ ( x∈ZZ, k∈ZZ).`
`d) |B| > 1 ⇔ |B| - 1 > 0 ⇔ |\frac{x-3}{6}| - 1 > 0 ⇔ |\frac{x-3}{6}| - 6/6 > 0.`
+) Với `\frac{x-3}{6} \ge 0 ⇔ x \ge 3 ⇒ |\frac{x-3}{6}| = \frac{x-3}{6}.`
`⇒\frac{x-3}{6}- 6/6 > 0`
`⇔\frac{x-3-6}{6} > 0`
`⇔\frac{x-9}{6} > 0`
`⇔\frac{x-9}{6} > 0`
`⇔x-9>0`
`⇔x>0+9`
`⇔x>9.` ( TMĐK )
+) Với `\frac{x-3}{6} < 0 ⇔ x < 3 ⇒ |\frac{x-3}{6}| = -\frac{x-3}{6}.`
`⇒-\frac{x-3}{6}- 6/6 > 0`
`⇔\frac{-x+3-6}{6} > 0`
`⇔\frac{-x-3}{6} > 0`
`⇔\frac{-(x+3)}{6} > 0`
`⇔-(x+3)>0`
`⇔x+3<0`
`⇔x<-3.` ( TMĐK )
Vậy với `x>9, x<-3 ` thì `|B| > 1.`
