[tex]\sqrt{x+8-2\sqrt{x+7}}=2-\sqrt{x+1-\sqrt{x+7}} (1)[/tex]
Điều kiện:[tex]-7\leq x\leq \frac{7+\sqrt{41}}{2}[/tex]
[tex](1)< = > \sqrt{(x+7)-2\sqrt{x+7}+1}=2\sqrt{(x+7)-2\sqrt{x+7}+1+(\sqrt{x+7}-7)}[/tex]
<=>[tex]\sqrt{(\sqrt{x+7}-1)^2}=2-\sqrt{(\sqrt{x+7}-1)^2+(\sqrt{x+7}-7)}(2)[/tex]
Đặt [tex]\sqrt{x+7}-1=a(a\geq 1)[/tex]
ta có:
(2)<=>[tex]a=2-\sqrt{a^2+a-6}[/tex]
<=>[tex]2-a=\sqrt{a^2+a-6}[/tex]
<=>[tex]4-4a+a^2=a^2+a-6[/tex]
<=>[tex]a=2[/tex] =>x=2