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

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

Whakaoti mō x

$$x = \frac{\sqrt{3} + 3}{2} \approx 2.366025404$$

$$x=\frac{3-\sqrt{3}}{2}\approx 0.633974596$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$2x^{2}-5x-3+3\times 2=\left(x-3\right)\left(1-2x\right)$$

Whakareatia te 3 ki te 2, ka 6.

$$2x^{2}-5x-3+6=\left(x-3\right)\left(1-2x\right)$$

Tāpirihia te -3 ki te 6, ka 3.

$$2x^{2}-5x+3=\left(x-3\right)\left(1-2x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te 1-2x ka whakakotahi i ngā kupu rite.

$$2x^{2}-5x+3=7x-2x^{2}-3$$

Tangohia te 7x mai i ngā taha e rua.

$$2x^{2}-5x+3-7x=-2x^{2}-3$$

Pahekotia te -5x me -7x, ka -12x.

$$2x^{2}-12x+3=-2x^{2}-3$$

Me tāpiri te 2x^{2} ki ngā taha e rua.

$$2x^{2}-12x+3+2x^{2}=-3$$

Pahekotia te 2x^{2} me 2x^{2}, ka 4x^{2}.

$$4x^{2}-12x+3=-3$$

Me tāpiri te 3 ki ngā taha e rua.

$$4x^{2}-12x+3+3=0$$

Tāpirihia te 3 ki te 3, ka 6.

$$4x^{2}-12x+6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -12 mō b, me 6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 6}}{2\times 4}$$

Pūrua -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 6}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 6}}{2\times 4}$$

Whakareatia -16 ki te 6.

$$x=\frac{-\left(-12\right)±\sqrt{144-96}}{2\times 4}$$

Tāpiri 144 ki te -96.

$$x=\frac{-\left(-12\right)±\sqrt{48}}{2\times 4}$$

Tuhia te pūtakerua o te 48.

$$x=\frac{-\left(-12\right)±4\sqrt{3}}{2\times 4}$$

Ko te tauaro o -12 ko 12.

$$x=\frac{12±4\sqrt{3}}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{12±4\sqrt{3}}{8}$$

Nā, me whakaoti te whārite x=\frac{12±4\sqrt{3}}{8} ina he tāpiri te ±. Tāpiri 12 ki te 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+12}{8}$$

Whakawehe 12+4\sqrt{3} ki te 8.

$$x=\frac{\sqrt{3}+3}{2}$$

Nā, me whakaoti te whārite x=\frac{12±4\sqrt{3}}{8} ina he tango te ±. Tango 4\sqrt{3} mai i 12.

$$x=\frac{12-4\sqrt{3}}{8}$$

Whakawehe 12-4\sqrt{3} ki te 8.

$$x=\frac{3-\sqrt{3}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{3}+3}{2} x=\frac{3-\sqrt{3}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$2x^{2}-5x-3+3\times 2=\left(x-3\right)\left(1-2x\right)$$

Whakareatia te 3 ki te 2, ka 6.

$$2x^{2}-5x-3+6=\left(x-3\right)\left(1-2x\right)$$

Tāpirihia te -3 ki te 6, ka 3.

$$2x^{2}-5x+3=\left(x-3\right)\left(1-2x\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x-3 ki te 1-2x ka whakakotahi i ngā kupu rite.

$$2x^{2}-5x+3=7x-2x^{2}-3$$

Tangohia te 7x mai i ngā taha e rua.

$$2x^{2}-5x+3-7x=-2x^{2}-3$$

Pahekotia te -5x me -7x, ka -12x.

$$2x^{2}-12x+3=-2x^{2}-3$$

Me tāpiri te 2x^{2} ki ngā taha e rua.

$$2x^{2}-12x+3+2x^{2}=-3$$

Pahekotia te 2x^{2} me 2x^{2}, ka 4x^{2}.

$$4x^{2}-12x+3=-3$$

Tangohia te 3 mai i ngā taha e rua.

$$4x^{2}-12x=-3-3$$

Tangohia te 3 i te -3, ka -6.

$$4x^{2}-12x=-6$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-12x}{4}=-\frac{6}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

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

Whakawehe -12 ki te 4.

$$x^{2}-3x=-\frac{6}{4}$$

Whakahekea te hautanga \frac{-6}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}-3x=-\frac{3}{2}$$

Whakawehea te -3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{2}. Nā, tāpiria te pūrua o te -\frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=-\frac{3}{2}+\frac{9}{4}$$

Tāpiri -\frac{3}{2} ki te \frac{9}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}-3x+\frac{9}{4}=\frac{3}{4}$$

Tauwehea x^{2}-3x+\frac{9}{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}.

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

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

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

Whakarūnātia.

$$x-\frac{3}{2}=\frac{\sqrt{3}}{2} x-\frac{3}{2}=-\frac{\sqrt{3}}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{3}+3}{2} x=\frac{3-\sqrt{3}}{2}$$