Lời giải:
ĐKXĐ: $x\geq 0; x\neq 25$
Gọi biểu thức là $A$
\(A=\frac{\sqrt{x}(\sqrt{x}+5)}{(\sqrt{x}-5)(\sqrt{x}+5)}-\frac{10\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}-\frac{5(\sqrt{x}-5)}{(\sqrt{x}+5)(\sqrt{x}-5)}\)
\(=\frac{\sqrt{x}(\sqrt{x}+5)-10\sqrt{x}-5(\sqrt{x}-5)}{(\sqrt{x}-5)(\sqrt{x}+5)}=\frac{x-10\sqrt{x}+25}{(\sqrt{x}-5)(\sqrt{x}+5)}\)
\(=\frac{(\sqrt{x}-5)^2}{(\sqrt{x}-5)(\sqrt{x}+5)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)