Giải thích các bước giải:
$1/ \frac{\sqrt[3]{x^3+3x^2+3x+4}}{\sqrt[]{x.|x|+2} }$
Đk: $\sqrt[]{x.|x|+2} >0$
$⇔x.|x| >-2$
$⇔x >-2$ (Vì: $|x| ≥0$)
$⇒$TXD: $x>-2$
$2/ \frac{\sqrt[4]{x^3+6x^2+12x+9}}{|x+5|-|x+3| }$
Đk: $\left \{ {{x^3+6x^2+12x+9 ≥0} \atop {|x+5|-|x+3| \neq 0}} \right.$
$⇔\left \{ {{x≥-3} \atop {|x+5| \neq |x+3|}} \right.$
$⇔\left \{ {{x≥-3} \atop {(x+5)^2 \neq (x+3)^2}} \right.$
$⇔ \left \{ {{x≥-3} \atop {4x+16 \neq 0}} \right.$
$⇔ \left \{ {{x≥-3} \atop {x \neq -4}} \right.$
$⇒$TXD: $x≥-3$
$3/ \frac{|5x+3|}{\sqrt[]{\sqrt[]{x^2+1} }+x}$
Đk: $\left \{ {{x^2+1 ≥0 (LĐ)} \atop {\sqrt[]{\sqrt[]{x^2+1} }+x \neq 0 (LĐ) }} \right.$
$⇒$TXD: $x \in R$
