Điều kiện: `x>=5`
$\sqrt{x}+\sqrt{x+7}+\sqrt{x-5}=9$
$⇔\sqrt{x}+\sqrt{x+7}+\sqrt{x-5}-9=0$
$⇔(\sqrt{x}-3)+(\sqrt{x+7}-4)+(\sqrt{x-5}-2)=0$
$⇔\dfrac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}+3}+\dfrac{(\sqrt{x+7}-4)(\sqrt{x+7}+4)}{\sqrt{x+7}+4}+\dfrac{(\sqrt{x-5}-2)(\sqrt{x-5}+2)}{\sqrt{x-5}+2}=0$
$⇔\dfrac{x-9}{\sqrt{x}+3}+\dfrac{x-9}{\sqrt{x+7}+4}+\dfrac{x-9}{\sqrt{x-5}+2}=0$
`<=>(x-9)(1/(sqrtx+3)+1/(sqrt(x+7)+4)+1/(sqrt(x-5)+2))=0`
Vì `1/(sqrtx+3)+1/(sqrt(x+7)-4)+1/(sqrt(x-5)+2)>0`
`=>x-9=0`
$\Leftrightarrow x=9(t/m)$
Vậy phương trình có nghiệm duy nhất `x=9`
