Whakaoti i te x+1divx=5

$$x+1 \div x=5$$

Whakaoti mō x

$$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$$

$$x=\frac{5-\sqrt{21}}{2}\approx 0.208712153$$

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.

$$xx+1=5x$$

Whakareatia te x ki te x, ka x^{2}.

$$x^{2}+1=5x$$

Tangohia te 5x mai i ngā taha e rua.

$$x^{2}+1-5x=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x^{2}-5x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4}}{2}$$

Tāpiri 25 ki te -4.

$$x=\frac{-\left(-5\right)±\sqrt{21}}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{21}}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{21}}{2} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{21}.

$$x=\frac{\sqrt{21}+5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{21}}{2} ina he tango te ±. Tango \sqrt{21} mai i 5.

$$x=\frac{5-\sqrt{21}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$

Ngā Upane mō te Whakaoti i te 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.

$$xx+1=5x$$

Whakareatia te x ki te x, ka x^{2}.

$$x^{2}+1=5x$$

Tangohia te 5x mai i ngā taha e rua.

$$x^{2}+1-5x=0$$

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-5x=-1$$

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-1+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=-1+\frac{25}{4}$$

Tāpiri -1 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{21}{4}$$

Tauwehea x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{21}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{\sqrt{21}}{2} x-\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{21}+5}{2} x=\frac{5-\sqrt{21}}{2}$$