a: ĐKXĐ: x>=0; x<>9
Sửa đề: \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Thay \(x=11+6\sqrt{2}\) vào M,ta được:
\(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)
c: Để M<1 thì M-1<0
\(\Leftrightarrow\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>căn x-3<0
=>0<=x<9