Whakaoti i te x^2+1/4+1/x^2=3/2

$$\frac { x ^ { 2 } + 1 } { 4 } + \frac { 1 } { x ^ { 2 } } = \frac { 3 } { 2 }$$

Whakaoti mō x

$$x=-1$$

$$x=1$$

$$x=2$$

$$x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 4x^{2}, arā, te tauraro pātahi he tino iti rawa te kitea o 4,x^{2},2.

$$x^{2}\left(x^{2}+1\right)+4=6x^{2}$$

Whakamahia te āhuatanga tohatoha hei whakarea te x^{2} ki te x^{2}+1.

$$x^{4}+x^{2}+4=6x^{2}$$

Tangohia te 6x^{2} mai i ngā taha e rua.

$$x^{4}+x^{2}+4-6x^{2}=0$$

Pahekotia te x^{2} me -6x^{2}, ka -5x^{2}.

$$x^{4}-5x^{2}+4=0$$

Whakakapia te t mō te x^{2}.

$$t^{2}-5t+4=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 1 mō te a, te -5 mō te b, me te 4 mō te c i te ture pūrua.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 4}}{2}$$

Mahia ngā tātaitai.

$$t=\frac{5±3}{2}$$

Whakaotia te whārite t=\frac{5±3}{2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=4 t=1$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō ia t.

$$x=2 x=-2 x=1 x=-1$$