Giải thích các bước giải:
$y=\frac{5}{x^2-1}+\sqrt[]{3-x}$
Đk: $\left \{ {{x^2-1 \neq 0} \atop {3-x ≥0}} \right.$
$⇔\left \{ {{x \neq ±1 } \atop {x≤3}} \right.$
$⇒TXD: D_1=(-∞;3]$\{$±1$}
$y=\frac{5}{\sqrt[]{-3x-1}}+ \sqrt[]{4+x}$
Đk: $\left \{ {{-3x-1>0} \atop {x+4≥0}} \right.$
$⇔\left \{ {{x<\frac{-1}{3} } \atop {x ≥-4}} \right.$
$⇒TXD: D_2=[-4; -\frac{1}{3})$
$a, D_1 ∩ D_2=[-4; -\frac{1}{3})$\{$-1$}
$b, C_R(D_1) ∩ D_2$
Ta có: $C_R(D_1)=(3;+∞) ∨ ${$±1$}
$⇒ C_R(D_1) ∩ D_2=-1$