Question

Cho biểu thức: \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

a) Rút gọn P

b) Tìm x để P < \(\frac{1}{2}\)

c) Tìm GTNN của P

Answer 1

ĐKXĐ: \(x\ge0;x\ne9\)

\(P=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\left(\sqrt{x}-3\right)}\right)\)

\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)\(P=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+3}\)

b/ Để P<\(\frac{1}{2}\Leftrightarrow\frac{-3}{\sqrt{x}+3}< \frac{1}{2}\)

Có \(\sqrt{x}+3>0\forall x\Rightarrow\frac{-3}{\sqrt{x}+3}< 0< \frac{1}{2}\forall x\ne9;x\ge0\)

Answer 2

c/ để P nhỏ nhất\(\Leftrightarrow\sqrt{x}+3max\)

Có \(\sqrt{x}+3\ge3\forall tmĐKXĐ\)

"="\(\Leftrightarrow x=0\)

Answer 3

GTNN của P là -1