Whakaoti i te 3x/2x-2+x/1-x-9/2x+2=0

$$\frac{ 3x }{ 2x-2 } + \frac{ x }{ 1-x } - \frac{ 9 }{ 2x+2 } =0$$

Whakaoti mō x

$$x=\sqrt{7}+4\approx 6.645751311$$

$$x=4-\sqrt{7}\approx 1.354248689$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 2\left(x-1\right)\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x-2,1-x,2x+2.

$$\left(x+1\right)\times 3x+\left(-2-2x\right)x-\left(x-1\right)\times 9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 3.

$$\left(3x+3\right)x+\left(-2-2x\right)x-\left(x-1\right)\times 9=0$$

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

$$3x^{2}+3x+\left(-2-2x\right)x-\left(x-1\right)\times 9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -2-2x ki te x.

$$3x^{2}+3x-2x-2x^{2}-\left(x-1\right)\times 9=0$$

Pahekotia te 3x me -2x, ka x.

$$3x^{2}+x-2x^{2}-\left(x-1\right)\times 9=0$$

Pahekotia te 3x^{2} me -2x^{2}, ka x^{2}.

$$x^{2}+x-\left(x-1\right)\times 9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te 9.

$$x^{2}+x-\left(9x-9\right)=0$$

Hei kimi i te tauaro o 9x-9, kimihia te tauaro o ia taurangi.

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

Pahekotia te x me -9x, ka -8x.

$$x^{2}-8x+9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -8 mō b, me 9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 9}}{2}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 9}}{2}$$

Whakareatia -4 ki te 9.

$$x=\frac{-\left(-8\right)±\sqrt{64-36}}{2}$$

Tāpiri 64 ki te -36.

$$x=\frac{-\left(-8\right)±\sqrt{28}}{2}$$

Tuhia te pūtakerua o te 28.

$$x=\frac{-\left(-8\right)±2\sqrt{7}}{2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±2\sqrt{7}}{2}$$

Nā, me whakaoti te whārite x=\frac{8±2\sqrt{7}}{2} ina he tāpiri te ±. Tāpiri 8 ki te 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+8}{2}$$

Whakawehe 8+2\sqrt{7} ki te 2.

$$x=\sqrt{7}+4$$

Nā, me whakaoti te whārite x=\frac{8±2\sqrt{7}}{2} ina he tango te ±. Tango 2\sqrt{7} mai i 8.

$$x=\frac{8-2\sqrt{7}}{2}$$

Whakawehe 8-2\sqrt{7} ki te 2.

$$x=4-\sqrt{7}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{7}+4 x=4-\sqrt{7}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -1,1 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 2\left(x-1\right)\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x-2,1-x,2x+2.

$$\left(x+1\right)\times 3x+\left(-2-2x\right)x-\left(x-1\right)\times 9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x+1 ki te 3.

$$\left(3x+3\right)x+\left(-2-2x\right)x-\left(x-1\right)\times 9=0$$

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

$$3x^{2}+3x+\left(-2-2x\right)x-\left(x-1\right)\times 9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -2-2x ki te x.

$$3x^{2}+3x-2x-2x^{2}-\left(x-1\right)\times 9=0$$

Pahekotia te 3x me -2x, ka x.

$$3x^{2}+x-2x^{2}-\left(x-1\right)\times 9=0$$

Pahekotia te 3x^{2} me -2x^{2}, ka x^{2}.

$$x^{2}+x-\left(x-1\right)\times 9=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-1 ki te 9.

$$x^{2}+x-\left(9x-9\right)=0$$

Hei kimi i te tauaro o 9x-9, kimihia te tauaro o ia taurangi.

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

Pahekotia te x me -9x, ka -8x.

$$x^{2}-8x+9=0$$

Tangohia te 9 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-8x=-9$$

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=-9+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=-9+16$$

Tāpiri -9 ki te 16.

$$x^{2}-8x+16=7$$

Tauwehea x^{2}-8x+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}.

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

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

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{7}$$

Whakarūnātia.

$$x-4=\sqrt{7} x-4=-\sqrt{7}$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=\sqrt{7}+4 x=4-\sqrt{7}$$