Lời giải:
ĐK: $x\geq 0$
a)
\(A=\frac{\sqrt{x}(\sqrt{x^3}+1)}{x-\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}\)
\(=\sqrt{x}(\sqrt{x}+1)-\sqrt{x}(\sqrt{x}-1)=2\sqrt{x}\)
b)
\(x=29-12\sqrt{5}=20+9-2\sqrt{20.9}=(\sqrt{20}-\sqrt{9})^2\Rightarrow \sqrt{x}=\sqrt{20}-3\)
Do đó: $A=2\sqrt{x}=2(\sqrt{20}-3)$
c)
$x+A=m\Leftrightarrow x+2\sqrt{x}=m$ (đề không đủ ý)