Whakaoti i te x^2-4x-1/(x^2+1)^2=0

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

Whakaoti mō x

$$x=\sqrt{5}+2\approx 4.236067977$$

$$x=2-\sqrt{5}\approx -0.236067977$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te \left(x^{2}+1\right)^{2}.

$$x^{2}-4x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -4 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)}}{2}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4}}{2}$$

Tāpiri 16 ki te 4.

$$x=\frac{-\left(-4\right)±\sqrt{20}}{2}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-\left(-4\right)±2\sqrt{5}}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+4}{2}$$

Whakawehe 4+2\sqrt{5} ki te 2.

$$x=\sqrt{5}+2$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{5}}{2} ina he tango te ±. Tango 2\sqrt{5} mai i 4.

$$x=\frac{4-2\sqrt{5}}{2}$$

Whakawehe 4-2\sqrt{5} ki te 2.

$$x=2-\sqrt{5}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{5}+2 x=2-\sqrt{5}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te \left(x^{2}+1\right)^{2}.

$$x^{2}-4x-1=0$$

Me tāpiri te 1 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x^{2}-4x=1$$

Whakawehea te -4, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -2. Nā, tāpiria te pūrua o te -2 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūrua -2.

$$x^{2}-4x+4=1+4$$

Tāpiri 1 ki te 4.

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

Tauwehea x^{2}-4x+4. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=5$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{5}$$

Whakarūnātia.

$$x-2=\sqrt{5} x-2=-\sqrt{5}$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=\sqrt{5}+2 x=2-\sqrt{5}$$