Whakaoti i te 4(1+1/x)=x^2-1

$$4 ( 1 + \frac { 1 } { x } ) = x ^ { 2 } - 1$$

Whakaoti mō x

$$x=\frac{1-\sqrt{17}}{2}\approx -1.561552813$$

$$x=-1$$

$$x = \frac{\sqrt{17} + 1}{2} \approx 2.561552813$$

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. Whakareatia ngā taha e rua o te whārite ki te x.

$$4\left(1+\frac{1}{x}\right)x=xx^{2}+x\left(-1\right)$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 1 me te 2 kia riro ai te 3.

$$4\left(1+\frac{1}{x}\right)x=x^{3}+x\left(-1\right)$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 1 ki te \frac{x}{x}.

$$4\left(\frac{x}{x}+\frac{1}{x}\right)x=x^{3}+x\left(-1\right)$$

Tā te mea he rite te tauraro o \frac{x}{x} me \frac{1}{x}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$4\times \frac{x+1}{x}x=x^{3}+x\left(-1\right)$$

Tuhia te 4\times \frac{x+1}{x} hei hautanga kotahi.

$$\frac{4\left(x+1\right)}{x}x=x^{3}+x\left(-1\right)$$

Tuhia te \frac{4\left(x+1\right)}{x}x hei hautanga kotahi.

$$\frac{4\left(x+1\right)x}{x}=x^{3}+x\left(-1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x+1.

$$\frac{\left(4x+4\right)x}{x}=x^{3}+x\left(-1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4x+4 ki te x.

$$\frac{4x^{2}+4x}{x}=x^{3}+x\left(-1\right)$$

Tangohia te x^{3} mai i ngā taha e rua.

$$\frac{4x^{2}+4x}{x}-x^{3}=x\left(-1\right)$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x^{3} ki te \frac{x}{x}.

$$\frac{4x^{2}+4x}{x}-\frac{x^{3}x}{x}=x\left(-1\right)$$

Tā te mea he rite te tauraro o \frac{4x^{2}+4x}{x} me \frac{x^{3}x}{x}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{4x^{2}+4x-x^{3}x}{x}=x\left(-1\right)$$

Mahia ngā whakarea i roto o 4x^{2}+4x-x^{3}x.

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

Tangohia te x\left(-1\right) mai i ngā taha e rua.

$$\frac{4x^{2}+4x-x^{4}}{x}-x\left(-1\right)=0$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia x\left(-1\right) ki te \frac{x}{x}.

$$\frac{4x^{2}+4x-x^{4}}{x}-\frac{x\left(-1\right)x}{x}=0$$

Tā te mea he rite te tauraro o \frac{4x^{2}+4x-x^{4}}{x} me \frac{x\left(-1\right)x}{x}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{4x^{2}+4x-x^{4}-x\left(-1\right)x}{x}=0$$

Mahia ngā whakarea i roto o 4x^{2}+4x-x^{4}-x\left(-1\right)x.

$$\frac{4x^{2}+4x-x^{4}+x^{2}}{x}=0$$

Whakakotahitia ngā kupu rite i 4x^{2}+4x-x^{4}+x^{2}.

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

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te x.

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

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.

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

Mahia ngā tātaitai.

$$t=\frac{-5±\sqrt{41}}{-2}$$

Whakaotia te whārite t=\frac{-5±\sqrt{41}}{-2} ina he tōrunga te ±, ina he tōraro te ±.

$$t=\frac{5-\sqrt{41}}{2} t=\frac{\sqrt{41}+5}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{41}+10}}{2} x=-\frac{\sqrt{2\sqrt{41}+10}}{2}$$