Đáp án:
a) $N=\dfrac{4\sqrt x}{3(x-\sqrt x+1)}$
b) $x\in\left\{\dfrac{1}{4};4\right\}$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0$
a) $N=\left(\dfrac{x+2}{x\sqrt x+1}-\dfrac{1}{\sqrt x+1}\right).\dfrac{4\sqrt x}{3}$
$=\left(\dfrac{x+2}{\sqrt x^3+1^3}-\dfrac{1}{\sqrt x+1}\right).\dfrac{4\sqrt x}{3}$
$=\left(\dfrac{x+2}{(\sqrt x+1)(x-\sqrt x+1)}-\dfrac{x-\sqrt x+1}{(\sqrt x+1)(x-\sqrt x+1)}\right).\dfrac{4\sqrt x}{3}$
$=\left(\dfrac{x+2-x+\sqrt x-1}{(\sqrt x+1)(x-\sqrt x+1)}\right).\dfrac{4\sqrt x}{3}$
$=\left(\dfrac{\sqrt x+1}{(\sqrt x+1)(x-\sqrt x+1)}\right).\dfrac{4\sqrt x}{3}$
$=\dfrac{1}{x-\sqrt x+1}.\dfrac{4\sqrt x}{3}$
$=\dfrac{4\sqrt x}{3(x-\sqrt x+1)}$
b) $N=\dfrac{8}{9}$
$⇒\dfrac{4\sqrt x}{3(x-\sqrt x+1)}=\dfrac{8}{9}$
$⇒\dfrac{\sqrt x}{x-\sqrt x+1}=\dfrac{2}{3}$
$⇒3\sqrt x=2x-2\sqrt x+2$
$⇒2x-5\sqrt x+2=0$
$⇒(2\sqrt x-1)(\sqrt x-2)=0$
$⇒\left[ \begin{array}{l}2\sqrt x-1=0\\\sqrt x-2=0\end{array} \right.⇒\left[ \begin{array}{l}\sqrt x=\dfrac{1}{2}\\\sqrt x=2\end{array} \right.⇒\left[ \begin{array}{l}x=\dfrac{1}{4}\\x=4\end{array} \right.$
Vậy $N=\dfrac{8}{9}$ khi $x\in\left\{\dfrac{1}{4};4\right\}$.
