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C = \(=\frac{x+3+2\text{ }\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x+3\right)\left(x-3\right)}}=\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)
Với x<-3 ta có:
\(x+3+2\sqrt{x^2-9}=\sqrt{-\left(x+3\right)}.\sqrt{-\left(x+3\right)}+2\sqrt{-\left(x+3\right)}.\sqrt{3-x}\)
\(=\sqrt{-\left(x+3\right)}.\left(\sqrt{-\left(x+3\right)}+2\sqrt{3-x}\right)\)
\(6-2x+\sqrt{x^2-9}=\sqrt{3-x}\left(2\sqrt{3-x}+\sqrt{-\left(x+3\right)}\right)\)
Từ đó suy ra \(M=\frac{\sqrt{-\left(x+3\right)}}{\sqrt{3-x}}hayM=\sqrt{\frac{\left(x+3\right)}{\left(x-3\right)}}\)
\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\frac{\left(\sqrt{x+3}\right)^2+2\sqrt{x+3}\sqrt{x-3}}{2.\left(\sqrt{x-3}\right)^2+\sqrt{x+3}\sqrt{x-3}}\)
\(=\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)
\(=\frac{\sqrt{x^2-9}}{x-3}\)
\(A=\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\\ ĐKXĐ:x\ne3\\ A=\frac{x+3+2\sqrt{\left(x+3\right)\left(x-3\right)}}{2\left(x-3\right)+\sqrt{\left(x+3\right)\left(x-3\right)}}\\ =\frac{\sqrt{x+3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)