a: \(=\dfrac{-13\sqrt{x}+3+2x+6\sqrt{x}+x-2\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x-9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)
b: Để A=1/2 thì \(\dfrac{3\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\)
=>\(6\sqrt{x}=\sqrt{x}+3\)
=>5 căn x=3
=>x=9/25
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$
\(A=\frac{3-13\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}+\frac{2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\frac{(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\)
\(=\frac{3-13\sqrt{x}+2\sqrt{x}(\sqrt{x}+3)+(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\)
\(=\frac{3x-9\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3\sqrt{x}}{\sqrt{x}+3}\)
b. Để $A=\frac{3\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}$
$\Leftrightarrow 6\sqrt{x}=\sqrt{x}+3$
$\Leftrightarrow 5\sqrt{x}=3\Leftrightarrow x=\frac{9}{25}$ (tm)
