cho `(x^n-x^(-n))/(x^(n)+x^(-n))=m` ; n ∈ N*, tính `P=(x^(2n)-x^(-2n))/(x^(2n)+x^(-2n))` theo `m`

0 bình luận về “cho `(x^n-x^(-n))/(x^(n)+x^(-n))=m` ; n ∈ N*, tính `P=(x^(2n)-x^(-2n))/(x^(2n)+x^(-2n))` theo `m`”

1. Ta có:

   *`m={x^n -x^{-n}}/{x^n +x^{-n}}`

   `=>m^2 ={(x^n -x^{-n})^2}/{(x^n +x^{-n})^2}`

   `<=>m^2={x^{2n}-2x^n .x^{-n}+x^{-2n}}/{x^{2n}+2x^n . x^{-n}+x^{-2n}}`

   `<=>m^2={x^{2n}-2+x^{-2n}}/{x^{2n}+2+x^{-2n}}`

   `=>x^{2n}-2+x^{-2n}=m^2 .(x^{2n}+2+x^{-2n})`

   `<=>(m^2 -1)(x^{2n}+x^{-2n})=-2-2m^2`

   `=>x^{2n}+x^{-2n}={-2-2m^2}/{m^2 -1}` `(m≠±1)`

   *`m={x^n -x^{-n}}/{x^n +x^{-n}}={(x^n -x^{-n})(x^n +x^{-n})}/(x^n +x^{-n})^2`

   `<=>m={x^{2n} -x^{-2n}}/{x^{2n}+2.x^n . x^{-n}+x^{-2n}}`

   `<=>m={x^{2n}-x^{-2n}}/{x^{2n}+2+x^{-2n}}`

   `<=>x^{2n} -x^{-2n}=m(x^{2n}+2+x^{-2n})`

   `<=>{x^{2n} -x^{-2n}}/{x^{2n} +x^{-2n}}={m(x^{2n}+2+x^{-2n})}/{x^{2n} +x^{-2n}}`

   `<=>P=m+{2m}/{x^{2n} +x^{-2n}}`

   `⇔P=m+{2m}/{{-2-2m^2}/{m^2 -1}}`

   `<=>P=m+{2m(m^2 -1)}/{-2-2m^2}`

   `<=>P={-2m-2m^3+2m^3 -2m}/{-2-2m^2}`

   `<=>P={-4m}/{-2(1+m^2 )}={2m}/{m^2 +1}`

   Vậy `P={2m}/{m^2 +1}`

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