Đáp án:
Thay `x=2+\sqrt{3}` vào `M` ta có:
`M=((2+\sqrt{3}-1).\sqrt{3})/(\sqrt{(2+\sqrt{3})^2-(2+\sqrt{3})+1)`
`=((1+\sqrt{3}).\sqrt{3})/(\sqrt{4+4\sqrt{3}+3-2-\sqrt{3}+1`
`=((1+\sqrt{3}).\sqrt{3})/(\sqrt{6+3\sqrt{3}`
`=((1+\sqrt{3}).\sqrt{3})/(\sqrt{3.(2+\sqrt{3})`
`=((1+\sqrt{3}).\sqrt{3})/(\sqrt{3}.\sqrt{2+\sqrt{3})`
`=(1+\sqrt{3})/(\sqrt{2+\sqrt{3})`
`=((1+\sqrt{3}).(\sqrt{2-\sqrt{3}}))/(\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}))`
`=(\sqrt{(2-\sqrt{3}).(1+\sqrt{3})^2})/(\sqrt{2^2-(\sqrt{3})^2)`
`=(\sqrt{(2-\sqrt{3}).(4+2\sqrt{3})})/(1)`
`=\sqrt{2.(2-\sqrt{3}).(2+\sqrt{3})}`
`=\sqrt{2.(2^2-(\sqrt{3})^2)}=\sqrt{2.1}=\sqrt{2}`
Vậy `M=\sqrt{2}` khi `x=2+\sqrt{3}`
