Question

\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}\)

a, rút gọn A

b, tìm giá trị của x để \(A=\frac{1}{3}\)

c, tìm gtln của A

Answer 1

a, ĐKXĐ: \(x\ge0;x\ne9\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3}{\sqrt{x+3}}\)

b, \(A=\frac{1}{3}\Leftrightarrow\frac{3}{\sqrt{x+3}}=\frac{1}{3}\Leftrightarrow\sqrt{x}+3=9\Leftrightarrow x=36\)

c, \(\sqrt{x}+3\ge3>0\Rightarrow\frac{3}{\sqrt{x}+3}\le1\)

\(\Rightarrow Max_A=1\Leftrightarrow x=0\)