Đáp án + giải thích các bước giải:
a) Đặt `A=\sqrt{2+\sqrt{3}}`
`->\sqrt{2}A=\sqrt{4+2\sqrt{3}}`
`=\sqrt{3+2\sqrt{3}+1}`
`=\sqrt{(\sqrt{3}+1)^2}`
`=|\sqrt{3}+1|`
`=\sqrt{3}+1`
`->A=(\sqrt{3}+1)/\sqrt{2}=(\sqrt{6}+\sqrt{2})/2`
b) Đặt `B=\sqrt{2-\sqrt{3}}`
`->\sqrt{2}B=\sqrt{4-2\sqrt{3}}`
`=\sqrt{3-2\sqrt{3}+1}`
`=\sqrt{(\sqrt{3}-1)^2}`
`=|\sqrt{3}-1|`
`=\sqrt{3}-1`
`->B=(\sqrt{3}-1)/\sqrt{2}=(\sqrt{6}-\sqrt{2})/\sqrt{2}`
c) `\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}`
`=\sqrt{2-2\sqrt{2}+1}-\sqrt{2+2\sqrt{2}+1}`
`=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{2}+1)^2}`
`=|\sqrt{2}-1|-|\sqrt{2}+1|`
`=\sqrt{2}-1-(\sqrt{2}+1)`
`=\sqrt{2}-1-\sqrt{2}-1`
`=-2`
d) `\sqrt{3-\sqrt{5}}(\sqrt{10}-\sqrt{2})(3+\sqrt{5})`
`=\sqrt{6-2\sqrt{5}}(\sqrt{5}-1)(3+\sqrt{5})`
`=\sqrt{5-2\sqrt{5}+1}(\sqrt{5}-1)(3+\sqrt{5})`
`=\sqrt{(\sqrt{5}-1)^2}(\sqrt{5}-1)(3+\sqrt{5})`
`=|\sqrt{5}-1|(\sqrt{5}-1)(3+\sqrt{5})`
`=(\sqrt{5}-1)^2(3+\sqrt{5})`
`=(5-2\sqrt{5}+1)(3+\sqrt{5})`
`=(6-2\sqrt{5})(3+\sqrt{5})`
`=2(3-\sqrt{5})(3+\sqrt{5})`
`=2(9-5)`
`=8`
