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

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

Whakaoti mō x

$$x = \frac{\sqrt{41} + 7}{2} \approx 6.701562119$$

$$x=\frac{7-\sqrt{41}}{2}\approx 0.298437881$$

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

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

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

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

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

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

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

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

Pahekotia te -5x me -2x, ka -7x.

$$3x^{2}-7x+3-3=2\left(x-1\right)\left(x+1\right)$$

Tangohia te 3 i te 3, ka 0.

$$3x^{2}-7x=2\left(x-1\right)\left(x+1\right)$$

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

$$3x^{2}-7x=\left(2x-2\right)\left(x+1\right)$$

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

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

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

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

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

$$x^{2}-7x=-2$$

Me tāpiri te 2 ki ngā taha e rua.

$$x^{2}-7x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -7 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2}}{2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8}}{2}$$

Tāpiri 49 ki te -8.

$$x=\frac{-\left(-7\right)±\sqrt{41}}{2}$$

Ko te tauaro o -7 ko 7.

$$x=\frac{7±\sqrt{41}}{2}$$

Nā, me whakaoti te whārite x=\frac{7±\sqrt{41}}{2} ina he tāpiri te ±. Tāpiri 7 ki te \sqrt{41}.

$$x=\frac{\sqrt{41}+7}{2}$$

Nā, me whakaoti te whārite x=\frac{7±\sqrt{41}}{2} ina he tango te ±. Tango \sqrt{41} mai i 7.

$$x=\frac{7-\sqrt{41}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{41}+7}{2} x=\frac{7-\sqrt{41}}{2}$$

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

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

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

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

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

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

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

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

Pahekotia te -5x me -2x, ka -7x.

$$3x^{2}-7x+3-3=2\left(x-1\right)\left(x+1\right)$$

Tangohia te 3 i te 3, ka 0.

$$3x^{2}-7x=2\left(x-1\right)\left(x+1\right)$$

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

$$3x^{2}-7x=\left(2x-2\right)\left(x+1\right)$$

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

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

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

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

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

$$x^{2}-7x=-2$$

Whakawehea te -7, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{7}{2}. Nā, tāpiria te pūrua o te -\frac{7}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{7}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-7x+\frac{49}{4}=-2+\frac{49}{4}$$

Tāpiri -2 ki te \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{41}{4}$$

Tauwehea x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{41}{4}$$

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

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Whakarūnātia.

$$x-\frac{7}{2}=\frac{\sqrt{41}}{2} x-\frac{7}{2}=-\frac{\sqrt{41}}{2}$$

Me tāpiri \frac{7}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{41}+7}{2} x=\frac{7-\sqrt{41}}{2}$$