Whakaoti i te 1/3+1/x=1/2+x-1/6x

$$\frac { 1 } { 3 } + \frac { 1 } { x } = \frac { 1 } { 2 + x } - \frac { 1 } { 6 x }$$

Whakaoti mō x (complex solution)

$$x=\frac{-5+\sqrt{87}i}{4}\approx -1.25+2.331844763i$$

$$x=\frac{-\sqrt{87}i-5}{4}\approx -1.25-2.331844763i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 6x\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 3,x,2+x,6x.

$$6x\left(x+2\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te x+2.

$$\left(6x^{2}+12x\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 6x^{2}+12x ki te \frac{1}{3}.

$$2x^{2}+4x+6x+12=6x-\left(x+2\right)$$

Pahekotia te 4x me 6x, ka 10x.

$$2x^{2}+10x+12=6x-\left(x+2\right)$$

Hei kimi i te tauaro o x+2, kimihia te tauaro o ia taurangi.

$$2x^{2}+10x+12=6x-x-2$$

Pahekotia te 6x me -x, ka 5x.

$$2x^{2}+10x+12=5x-2$$

Tangohia te 5x mai i ngā taha e rua.

$$2x^{2}+10x+12-5x=-2$$

Pahekotia te 10x me -5x, ka 5x.

$$2x^{2}+5x+12=-2$$

Me tāpiri te 2 ki ngā taha e rua.

$$2x^{2}+5x+12+2=0$$

Tāpirihia te 12 ki te 2, ka 14.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 5 mō b, me 14 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 14}}{2\times 2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4\times 2\times 14}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-5±\sqrt{25-8\times 14}}{2\times 2}$$

Whakareatia -8 ki te 14.

$$x=\frac{-5±\sqrt{25-112}}{2\times 2}$$

Tāpiri 25 ki te -112.

$$x=\frac{-5±\sqrt{-87}}{2\times 2}$$

Tuhia te pūtakerua o te -87.

$$x=\frac{-5±\sqrt{87}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-5±\sqrt{87}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{87}i}{4} ina he tāpiri te ±. Tāpiri -5 ki te i\sqrt{87}.

$$x=\frac{-5+\sqrt{87}i}{4}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{87}i}{4} ina he tango te ±. Tango i\sqrt{87} mai i -5.

$$x=\frac{-\sqrt{87}i-5}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{-5+\sqrt{87}i}{4} x=\frac{-\sqrt{87}i-5}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 6x\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 3,x,2+x,6x.

$$6x\left(x+2\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te x+2.

$$\left(6x^{2}+12x\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 6x^{2}+12x ki te \frac{1}{3}.

$$2x^{2}+4x+6x+12=6x-\left(x+2\right)$$

Pahekotia te 4x me 6x, ka 10x.

$$2x^{2}+10x+12=6x-\left(x+2\right)$$

Hei kimi i te tauaro o x+2, kimihia te tauaro o ia taurangi.

$$2x^{2}+10x+12=6x-x-2$$

Pahekotia te 6x me -x, ka 5x.

$$2x^{2}+10x+12=5x-2$$

Tangohia te 5x mai i ngā taha e rua.

$$2x^{2}+10x+12-5x=-2$$

Pahekotia te 10x me -5x, ka 5x.

$$2x^{2}+5x+12=-2$$

Tangohia te 12 mai i ngā taha e rua.

$$2x^{2}+5x=-2-12$$

Tangohia te 12 i te -2, ka -14.

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

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+5x}{2}=-\frac{14}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}+\frac{5}{2}x=-\frac{14}{2}$$

Whakawehe -14 ki te 2.

$$x^{2}+\frac{5}{2}x=-7$$

Whakawehea te \frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{4}. Nā, tāpiria te pūrua o te \frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-7+\left(\frac{5}{4}\right)^{2}$$

Pūruatia \frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-7+\frac{25}{16}$$

Tāpiri -7 ki te \frac{25}{16}.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{87}{16}$$

Tauwehea x^{2}+\frac{5}{2}x+\frac{25}{16}. 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}.

$$\left(x+\frac{5}{4}\right)^{2}=-\frac{87}{16}$$

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

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{87}{16}}$$

Whakarūnātia.

$$x+\frac{5}{4}=\frac{\sqrt{87}i}{4} x+\frac{5}{4}=-\frac{\sqrt{87}i}{4}$$

Me tango \frac{5}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{-5+\sqrt{87}i}{4} x=\frac{-\sqrt{87}i-5}{4}$$