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

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

Whakaoti mō x

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4}\approx 0.153112887$$

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4}\approx -0.653112887$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2x-1 ki te x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te -1-2x ki te 2.

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

Pahekotia te -x me -4x, ka -5x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 2x-1.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te 6x-3 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$2x^{2}-5x-2=12x^{2}-3$$

Tangohia te 12x^{2} mai i ngā taha e rua.

$$2x^{2}-5x-2-12x^{2}=-3$$

Pahekotia te 2x^{2} me -12x^{2}, ka -10x^{2}.

$$-10x^{2}-5x-2=-3$$

Me tāpiri te 3 ki ngā taha e rua.

$$-10x^{2}-5x-2+3=0$$

Tāpirihia te -2 ki te 3, ka 1.

$$-10x^{2}-5x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -10 mō a, -5 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-10\right)}}{2\left(-10\right)}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-10\right)}}{2\left(-10\right)}$$

Whakareatia -4 ki te -10.

$$x=\frac{-\left(-5\right)±\sqrt{25+40}}{2\left(-10\right)}$$

Tāpiri 25 ki te 40.

$$x=\frac{-\left(-5\right)±\sqrt{65}}{2\left(-10\right)}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{65}}{2\left(-10\right)}$$

Whakareatia 2 ki te -10.

$$x=\frac{5±\sqrt{65}}{-20}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{65}}{-20} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{65}.

$$x=\frac{\sqrt{65}+5}{-20}$$

Whakawehe 5+\sqrt{65} ki te -20.

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{65}}{-20} ina he tango te ±. Tango \sqrt{65} mai i 5.

$$x=\frac{5-\sqrt{65}}{-20}$$

Whakawehe 5-\sqrt{65} ki te -20.

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Kua oti te whārite te whakatau.

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4} x=\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2x-1 ki te x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te -1-2x ki te 2.

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

Pahekotia te -x me -4x, ka -5x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 2x-1.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te 6x-3 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$2x^{2}-5x-2=12x^{2}-3$$

Tangohia te 12x^{2} mai i ngā taha e rua.

$$2x^{2}-5x-2-12x^{2}=-3$$

Pahekotia te 2x^{2} me -12x^{2}, ka -10x^{2}.

$$-10x^{2}-5x-2=-3$$

Me tāpiri te 2 ki ngā taha e rua.

$$-10x^{2}-5x=-3+2$$

Tāpirihia te -3 ki te 2, ka -1.

$$-10x^{2}-5x=-1$$

Whakawehea ngā taha e rua ki te -10.

$$\frac{-10x^{2}-5x}{-10}=-\frac{1}{-10}$$

Mā te whakawehe ki te -10 ka wetekia te whakareanga ki te -10.

$$x^{2}+\left(-\frac{5}{-10}\right)x=-\frac{1}{-10}$$

Whakahekea te hautanga \frac{-5}{-10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$x^{2}+\frac{1}{2}x=-\frac{1}{-10}$$

Whakawehe -1 ki te -10.

$$x^{2}+\frac{1}{2}x=\frac{1}{10}$$

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{1}{10}+\left(\frac{1}{4}\right)^{2}$$

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{10}+\frac{1}{16}$$

Tāpiri \frac{1}{10} ki te \frac{1}{16} 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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{13}{80}$$

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

$$\left(x+\frac{1}{4}\right)^{2}=\frac{13}{80}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{80}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{\sqrt{65}}{20} x+\frac{1}{4}=-\frac{\sqrt{65}}{20}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4} x=-\frac{\sqrt{65}}{20}-\frac{1}{4}$$