a/đkxđ \(x\ge0;x\ne9\)
b//đkxđ \(x\ge0;x\ne9\)
C=\(\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{\left(\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\right)}\right]\):\(\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
=\(\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=\(\frac{-3}{\sqrt{x}+3}\)
vậy C =\(\frac{-3}{\sqrt{x}+3}\) với \(x\ge0;x\ne9\)
c/để C<-1/2 thì:
\(\frac{-3}{\sqrt{x}+3}+\frac{1}{2}< 0\)\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\) (vì \(2\left(\sqrt{x}+3\right)>0\) do \(x\ge0;x\ne9\))
\(\Leftrightarrow x< 9\) kết hợp vs đkxđ
\(\Leftrightarrow0\le x< 9\)
vậy \(0\le x< 9\) thì C<-1/2
d/ ta có : \(x\ge0;x\ne9\)=> \(\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}+3\ge3\)\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}\ge-1\)
hay C\(\ge-1\) dấu "=" xảy ra khi: x=0(tm)
vậy min c=-1 tại x=0
