a.b. \(M=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\) \(\left(x\ne9,x\ge0\right)\)
\(M=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{36+8}{\sqrt{36}+1}=\dfrac{44}{7}\)
c. Mk nghĩ là GTLN :(
\(M=\dfrac{x+8}{\sqrt{x}+1}\le\dfrac{0+8}{0+1}=8\)
\("="\Leftrightarrow x=0\)
