Question

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

a, rút gọn P

b, tìm giá trị nhỏ nhất của P

Answer 1

\(x\ge0;x\ne9\)

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

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

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

Do \(-3< 0\Rightarrow P_{min}\) khi \(\sqrt{x}+3\) nhỏ nhất

Mà \(\sqrt{x}+3\ge3\Rightarrow P_{min}=\frac{-3}{3}=-1\) khi \(x=0\)