$ĐKXĐ:x\geq -2$
$Pt⇔$ Đặt $\sqrt[]{x+5}=a;\sqrt[]{x+2}=b$ ta có:
$1+a.b=a+b$
$⇔a+b-ab-1=0$
$⇔a(1-b)-(1-b)=0$
$⇔(1-b)(a-1)=0$
$⇔$\(\left[ \begin{array}{l}1-b=0\\a-1=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}\sqrt[]{x+2}=1\\\sqrt[]{x+5}=1\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x+2=1\\x+5=1\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-1(t/m\\x=-4(ko t/m)\end{array} \right.\)
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