Đáp án:
`sqrt{2/(2-sqrt{3})}-sqrt{2/(2+sqrt{3})}=2`
Giải thích các bước giải:
`b)`
`sqrt{2/(2-sqrt{3})}-sqrt{2/(2+sqrt{3})}=2`
Ta có:
`VT=sqrt{2/(2-sqrt{3})}-sqrt{2/(2+sqrt{3})}`
`VT=sqrt{(2.(2+sqrt{3}))/((2-sqrt{3})(2+sqrt{3}))}-sqrt{(2.(2-sqrt{3}))/((2+sqrt{3})(2-sqrt{3}))}`
`VT=sqrt{(4+2sqrt{3})/(4-3)}-sqrt{(4-2sqrt{3})/(4-3)}`
`VT=sqrt{(3+1+2.sqrt{3})/1-sqrt{3+1-2.sqrt{3})/1}`
`VT=sqrt{(sqrt{3})^2+2.sqrt{3}.1+1^2}-sqrt{(sqrt{3})^2-2.sqrt{3}.1+1^2}`
`VT=sqrt{(sqrt{3}+1)^2}-sqrt{(sqrt{3}-1)^2}`
`VT=|sqrt{3}+1|-|sqrt{3}-1|`
`VT=sqrt{3}+1-sqrt{3}+1` ( vì `sqrt{3}>sqrt{1}=1`)
`VT=2=VP(text{ĐPCM})`
Vậy `sqrt{2/(2-sqrt{3})}-sqrt{2/(2+sqrt{3})}=2`
