Đáp án: $x=0$
Giải thích các bước giải:
$ĐKXĐ: x\ge 0$
$\sqrt{x}-\sqrt{x+1}=\sqrt{x+4}-\sqrt{x+9}$
$\rightarrow (\sqrt{x}-\sqrt{x+1})^2=(\sqrt{x+4}-\sqrt{x+9})^2$
$\rightarrow x-2\sqrt{x}.\sqrt{x+1}+x+1=x+4-2\sqrt{x+4}.\sqrt{x+9}+x+9$
$\rightarrow 12=2\sqrt{(x+4)(x+9)}-2\sqrt{x(x+1)}$
$\rightarrow 6=\sqrt{x^2+13x+36}-\sqrt{x^2+x}$
$\rightarrow 6+\sqrt{x^2+x}=\sqrt{x^2+13x+36}$
$\rightarrow (6+\sqrt{x^2+x})^2=(\sqrt{x^2+13x+36})^2$
$\rightarrow 36+12\sqrt{x^2+x}+x^2+x=x^2+13x+36$
$\rightarrow x=\sqrt{x^2+x}$
$\rightarrow x-\sqrt{x}.\sqrt{x+1}=0$
$\rightarrow \sqrt{x}(\sqrt{x}-\sqrt{x+1})=0$
$\rightarrow x=0$
