Whakaoti i te 1/2x+2+1/4x-4=3

$$\frac { 1 } { 2 x + 2 } + \frac { 1 } { 4 x - 4 } = 3$$

Whakaoti mō x

$$x=\frac{\sqrt{537}}{24}+\frac{1}{8}\approx 1.090552519$$

$$x=-\frac{\sqrt{537}}{24}+\frac{1}{8}\approx -0.840552519$$

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 4\left(x-1\right)\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+2,4x-4.

$$2x-2+x+1=12\left(x-1\right)\left(x+1\right)$$

Pahekotia te 2x me x, ka 3x.

$$3x-2+1=12\left(x-1\right)\left(x+1\right)$$

Tāpirihia te -2 ki te 1, ka -1.

$$3x-1=12\left(x-1\right)\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 12 ki te x-1.

$$3x-1=\left(12x-12\right)\left(x+1\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 12x-12 ki te x+1 ka whakakotahi i ngā kupu rite.

$$3x-1=12x^{2}-12$$

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

$$3x-1-12x^{2}=-12$$

Me tāpiri te 12 ki ngā taha e rua.

$$3x-1-12x^{2}+12=0$$

Tāpirihia te -1 ki te 12, ka 11.

$$3x+11-12x^{2}=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$-12x^{2}+3x+11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -12 mō a, 3 mō b, me 11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-12\right)\times 11}}{2\left(-12\right)}$$

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-12\right)\times 11}}{2\left(-12\right)}$$

Whakareatia -4 ki te -12.

$$x=\frac{-3±\sqrt{9+48\times 11}}{2\left(-12\right)}$$

Whakareatia 48 ki te 11.

$$x=\frac{-3±\sqrt{9+528}}{2\left(-12\right)}$$

Tāpiri 9 ki te 528.

$$x=\frac{-3±\sqrt{537}}{2\left(-12\right)}$$

Whakareatia 2 ki te -12.

$$x=\frac{-3±\sqrt{537}}{-24}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{537}}{-24} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{537}.

$$x=\frac{\sqrt{537}-3}{-24}$$

Whakawehe -3+\sqrt{537} ki te -24.

$$x=-\frac{\sqrt{537}}{24}+\frac{1}{8}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{537}}{-24} ina he tango te ±. Tango \sqrt{537} mai i -3.

$$x=\frac{-\sqrt{537}-3}{-24}$$

Whakawehe -3-\sqrt{537} ki te -24.

$$x=\frac{\sqrt{537}}{24}+\frac{1}{8}$$

Kua oti te whārite te whakatau.

$$x=-\frac{\sqrt{537}}{24}+\frac{1}{8} x=\frac{\sqrt{537}}{24}+\frac{1}{8}$$

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 4\left(x-1\right)\left(x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x+2,4x-4.

$$2x-2+x+1=12\left(x-1\right)\left(x+1\right)$$

Pahekotia te 2x me x, ka 3x.

$$3x-2+1=12\left(x-1\right)\left(x+1\right)$$

Tāpirihia te -2 ki te 1, ka -1.

$$3x-1=12\left(x-1\right)\left(x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 12 ki te x-1.

$$3x-1=\left(12x-12\right)\left(x+1\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 12x-12 ki te x+1 ka whakakotahi i ngā kupu rite.

$$3x-1=12x^{2}-12$$

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

$$3x-1-12x^{2}=-12$$

Me tāpiri te 1 ki ngā taha e rua.

$$3x-12x^{2}=-12+1$$

Tāpirihia te -12 ki te 1, ka -11.

$$3x-12x^{2}=-11$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$-12x^{2}+3x=-11$$

Whakawehea ngā taha e rua ki te -12.

$$\frac{-12x^{2}+3x}{-12}=-\frac{11}{-12}$$

Mā te whakawehe ki te -12 ka wetekia te whakareanga ki te -12.

$$x^{2}+\frac{3}{-12}x=-\frac{11}{-12}$$

Whakahekea te hautanga \frac{3}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}-\frac{1}{4}x=-\frac{11}{-12}$$

Whakawehe -11 ki te -12.

$$x^{2}-\frac{1}{4}x=\frac{11}{12}$$

Whakawehea te -\frac{1}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{8}. Nā, tāpiria te pūrua o te -\frac{1}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{11}{12}+\left(-\frac{1}{8}\right)^{2}$$

Pūruatia -\frac{1}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{11}{12}+\frac{1}{64}$$

Tāpiri \frac{11}{12} ki te \frac{1}{64} 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.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{179}{192}$$

Tauwehea x^{2}-\frac{1}{4}x+\frac{1}{64}. 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}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{179}{192}$$

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

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{179}{192}}$$

Whakarūnātia.

$$x-\frac{1}{8}=\frac{\sqrt{537}}{24} x-\frac{1}{8}=-\frac{\sqrt{537}}{24}$$

Me tāpiri \frac{1}{8} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{537}}{24}+\frac{1}{8} x=-\frac{\sqrt{537}}{24}+\frac{1}{8}$$