`x ne -3; x ne 2`
$\text{Có: $\dfrac{x-4}{x-2}$ = 1 - $\dfrac{2}{x-2}$}$
$\text{Để B nguyên}$
`⇒x-2∈Ư(2)={±1;±2}`
`TH1: x-2=1⇔x=3(TM)`
`TH2: x-2=-1⇔x=1(TM)`
`TH3:x-2=2⇔x=4(TM)`
`TH4:x-2=-2⇔x=0(TM)`
$\text{Vậy để B nguyên thì x ∈ {0; 1; 3; 4}}$
`c, x^2 + 2x-3=0`
`⇔x^2+3x-x-3=0`
`⇔x(x+3)-(x+3)=0`
`⇔(x-1)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\x+3=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=1(TM)\\x=-3(KTM)\end{array} \right.\)
$\text{Thay x = 1 vào B}$
`⇒B=\frac{1-4}{1-2}=-3/-1=3`
$\text{Vậy tại x = 1 thì B = 3}$
