Whakaoti i te 3x+2+1/3x+2=7

$$3 x + 2 + \frac { 1 } { 3 x + 2 } = 7$$

Whakaoti mō x

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{2}{3} nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 3x+2.

$$3x\left(3x+2\right)+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te 3x+2.

$$9x^{2}+6x+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x+2 ki te 2.

$$9x^{2}+6x+6x+4+1=7\left(3x+2\right)$$

Pahekotia te 6x me 6x, ka 12x.

$$9x^{2}+12x+4+1=7\left(3x+2\right)$$

Tāpirihia te 4 ki te 1, ka 5.

$$9x^{2}+12x+5=7\left(3x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 7 ki te 3x+2.

$$9x^{2}+12x+5=21x+14$$

Tangohia te 21x mai i ngā taha e rua.

$$9x^{2}+12x+5-21x=14$$

Pahekotia te 12x me -21x, ka -9x.

$$9x^{2}-9x+5=14$$

Tangohia te 14 mai i ngā taha e rua.

$$9x^{2}-9x+5-14=0$$

Tangohia te 14 i te 5, ka -9.

$$9x^{2}-9x-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, -9 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 9\left(-9\right)}}{2\times 9}$$

Pūrua -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 9\left(-9\right)}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$x=\frac{-\left(-9\right)±\sqrt{81-36\left(-9\right)}}{2\times 9}$$

Whakareatia -36 ki te -9.

$$x=\frac{-\left(-9\right)±\sqrt{81+324}}{2\times 9}$$

Tāpiri 81 ki te 324.

$$x=\frac{-\left(-9\right)±\sqrt{405}}{2\times 9}$$

Tuhia te pūtakerua o te 405.

$$x=\frac{-\left(-9\right)±9\sqrt{5}}{2\times 9}$$

Ko te tauaro o -9 ko 9.

$$x=\frac{9±9\sqrt{5}}{2\times 9}$$

Whakareatia 2 ki te 9.

$$x=\frac{9±9\sqrt{5}}{18}$$

Nā, me whakaoti te whārite x=\frac{9±9\sqrt{5}}{18} ina he tāpiri te ±. Tāpiri 9 ki te 9\sqrt{5}.

$$x=\frac{9\sqrt{5}+9}{18}$$

Whakawehe 9+9\sqrt{5} ki te 18.

$$x=\frac{\sqrt{5}+1}{2}$$

Nā, me whakaoti te whārite x=\frac{9±9\sqrt{5}}{18} ina he tango te ±. Tango 9\sqrt{5} mai i 9.

$$x=\frac{9-9\sqrt{5}}{18}$$

Whakawehe 9-9\sqrt{5} ki te 18.

$$x=\frac{1-\sqrt{5}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{2}{3} nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 3x+2.

$$3x\left(3x+2\right)+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te 3x+2.

$$9x^{2}+6x+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x+2 ki te 2.

$$9x^{2}+6x+6x+4+1=7\left(3x+2\right)$$

Pahekotia te 6x me 6x, ka 12x.

$$9x^{2}+12x+4+1=7\left(3x+2\right)$$

Tāpirihia te 4 ki te 1, ka 5.

$$9x^{2}+12x+5=7\left(3x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 7 ki te 3x+2.

$$9x^{2}+12x+5=21x+14$$

Tangohia te 21x mai i ngā taha e rua.

$$9x^{2}+12x+5-21x=14$$

Pahekotia te 12x me -21x, ka -9x.

$$9x^{2}-9x+5=14$$

Tangohia te 5 mai i ngā taha e rua.

$$9x^{2}-9x=14-5$$

Tangohia te 5 i te 14, ka 9.

$$9x^{2}-9x=9$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9x^{2}-9x}{9}=\frac{9}{9}$$

Mā te whakawehe ki te 9 ka wetekia te whakareanga ki te 9.

$$x^{2}+\left(-\frac{9}{9}\right)x=\frac{9}{9}$$

Whakawehe -9 ki te 9.

$$x^{2}-x=\frac{9}{9}$$

Whakawehe 9 ki te 9.

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

Whakawehea te -1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{2}. Nā, tāpiria te pūrua o te -\frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri 1 ki te \frac{1}{4}.

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

Tauwehea x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

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

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

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$