a)
\(Q=\frac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{1}{\sqrt{x}+2}\\ =\frac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\frac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{1}{\sqrt{x}+2}\)
b)
\(Q=\frac{1}{\sqrt{x}+2}>\frac{1}{9}\\ \Leftrightarrow\sqrt{x}+2< 9\\ \Leftrightarrow\sqrt{x}< 7\Leftrightarrow x< 49\)
Vậy với \(0\le x\ne9\)thì Q>\(\frac{1}{9}\)
c) Để Q đạt GTLN thì \(\sqrt{x}+2\)đạt GTNN
Mà\(\sqrt{x}+2\ge2\forall x\)
Suy ra GTNN của \(\sqrt{x}+2=2\)khi x=0
Khi đó Max Q=\(\frac{1}{2}\)
d) a là giá trị nào vậy bạn ????
