Chứng minh `(2-sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))))/ (2-sqrt(2+sqrt(2+sqrt(2)))` `<1/3` - Yêu cầu giải chi tiết giúp mình ạ.
Chứng minh `(2-sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))))/ (2-sqrt(2+sqrt(2+sqrt(2)))` `<1/3` - Yêu cầu giải chi tiết giúp mình ạ.
Ta sử dụng phép biến đổi tương đương.
`(2-sqrt{2+sqrt{2+sqrt{2+sqrt2}}})/(2-sqrt{2+sqrt{2+sqrt2}})<1/3`
`<=>(6-3sqrt{2+sqrt{2+sqrt{2+sqrt2}}}-2+sqrt{2+sqrt{2+sqrt2}})/(3(2-sqrt{2+sqrt{2+sqrt2}}))<0`
`<=>(4-3sqrt{2+sqrt{2+sqrt{2+sqrt2}}}+sqrt{2+sqrt{2+sqrt2}})/(3(2-sqrt{2+sqrt{2+sqrt2}}))<0`
Vì `sqrt2<sqrt4`
`<=>sqrt{2+sqrt2}<sqrt{2+sqrt4}=2`
`<=>sqrt{2+sqrt{2+sqrt2}}<sqrt{2+2}=2`
`<=>2-sqrt{2+sqrt{2+sqrt2}}>2-2=0`
Vậy nhất thiết ta phải chứng minh `4-3sqrt{2+sqrt{2+sqrt{2+sqrt2}}}+sqrt{2+sqrt{2+sqrt2}}<0`.
Ta có:`sqrt2>sqrt{1,9881}`
`<=>sqrt2>1,41`
`<=>2+sqrt2>3,41`
`<=>sqrt{2+sqrt2}>sqrt{3,41}`
Mà `sqrt{3,41}>sqrt{3,3856}=1,84`
`<=>sqrt{2+sqrt2}>1,84`
`<=>2+sqrt{2+sqrt2}>3,84`
`<=>sqrt{2+sqrt{2+sqrt2}}>sqrt{3,84}`
Mà `sqrt{3,84}>sqrt{3,8025}=1,95`
`<=>2+sqrt{2+sqrt{2+sqrt2}}>3,95`
`<=>sqrt{2+sqrt{2+sqrt{2+sqrt2}}}>sqrt{3,95}`
Mà `sqrt{3,95}>sqrt{3,94975876}=1,9874`
`<=>sqrt{2+sqrt{2+sqrt{2+sqrt2}}}>1,9874`
`<=>3sqrt{2+sqrt{2+sqrt{2+sqrt2}}}>5,9622`
`sqrt2<sqrt{2,0164}=1,42`
`<=>2+sqrt2<3,42`
`<=>sqrt{2+sqrt2}<sqrt{3,42}`
Mà `sqrt{3,42}<sqrt{3,4225}=1,85`
`<=>2+sqrt{2+sqrt2}<3,85`
`<=>sqrt{2+sqrt{2+sqrt2}}<sqrt{3,85}`
Mà `sqrt{3,85}<sqrt{3,850032623}=1,96215`
`<=>sqrt{2+sqrt{2+sqrt2}}<1,96215<1,9622`
`<=>4+sqrt{2+sqrt{2+sqrt2}}<5,9622`
Mà `3sqrt{2+sqrt{2+sqrt{2+sqrt2}}}>5,9622`
`=>4+sqrt{2+sqrt{2+sqrt2}}-3sqrt{2+sqrt{2+sqrt{2+sqrt2}}}<0`
`=>` ta có điều phải chứng minh là `(2-sqrt{2+sqrt{2+sqrt{2+sqrt2}}})/(2-sqrt{2+sqrt{2+sqrt2}})<1/3`.
Đáp án:
`(2-sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))))/ (2-sqrt(2+sqrt(2+sqrt(2)))`
`= [(2 -sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) )(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) )]/[(2 – sqrt(2+sqrt(2+sqrt(2)) ))(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) )`
`= (2^2 – (sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))))^2)/[(2 – sqrt(2+sqrt(2+sqrt(2)) ))(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) )`
`= (4 – 2 – sqrt(2+sqrt(2+sqrt(2)) ) )/[(2 – sqrt(2+sqrt(2+sqrt(2)) ))(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) )`
`= (2 -sqrt(2+sqrt(2+sqrt(2)) ))/[(2 – sqrt(2+sqrt(2+sqrt(2)) ))(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) )`
`= 1/(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))))`
Do `sqrt(2+sqrt(2+ sqrt(2))) > -1 -> 2 + sqrt(2+sqrt(2+ sqrt(2))) > 2 + (-1) = 1`
`-> sqrt(2+sqrt(2+sqrt(2+ sqrt(2)))) > \sqrt{1} = 1`
`-> 1/(2 + sqrt(2+sqrt(2+sqrt(2+ sqrt(2))))) < 1/(2 + 1) = 1/3`
( ta có điều phải chứng minh )
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