Đáp án+Giải thích các bước giải:
`M=\frac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}(x≥2)`
`M=\frac{\sqrt{x-2+2\sqrt{(x-2)(x+2)}+x+2}}{\sqrt{(x-2)(x+2)}+x+2}`
`M=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}`
`M=\frac{|\sqrt{x-2}+\sqrt{x+2}|}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}`
`M=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}`
`M=\frac{1}{\sqrt{x+2}}`
Thay `x=2\sqrt{6}+3(tm)` vào `M` ta có:
`M=\frac{1}{\sqrt{2\sqrt{6}+3+2}}`
`M=\frac{1}{\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}}`
`M=\frac{1}{\sqrt{(\sqrt{3}+\sqrt{2})^2}`
`M=\frac{1}{|\sqrt{3}+\sqrt{2}|}`
`M=\frac{1}{\sqrt{3}+\sqrt{2}}`
`M=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}`
`M=\frac{\sqrt{3}-\sqrt{2}}{3-2}`
`M=\frac{\sqrt{3}-\sqrt{2}}{1}`
`M=\sqrt{3}-\sqrt{2}`
Vậy tại `x=2\sqrt{6}+3` thì `M=\sqrt{3}-\sqrt{2}`
