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

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

Whakaoti mō x

$$x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-3\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-3,x+2.

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

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

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

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

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

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

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

Pahekotia te 2x me -5x, ka -3x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 3.

$$3x^{2}-3x-3=3x+6$$

Tangohia te 3x mai i ngā taha e rua.

$$3x^{2}-3x-3-3x=6$$

Pahekotia te -3x me -3x, ka -6x.

$$3x^{2}-6x-3=6$$

Tangohia te 6 mai i ngā taha e rua.

$$3x^{2}-6x-3-6=0$$

Tangohia te 6 i te -3, ka -9.

$$3x^{2}-6x-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, -6 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 3\left(-9\right)}}{2\times 3}$$

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-9\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\left(-9\right)}}{2\times 3}$$

Whakareatia -12 ki te -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+108}}{2\times 3}$$

Tāpiri 36 ki te 108.

$$x=\frac{-\left(-6\right)±\sqrt{144}}{2\times 3}$$

Tuhia te pūtakerua o te 144.

$$x=\frac{-\left(-6\right)±12}{2\times 3}$$

Ko te tauaro o -6 ko 6.

$$x=\frac{6±12}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{6±12}{6}$$

Nā, me whakaoti te whārite x=\frac{6±12}{6} ina he tāpiri te ±. Tāpiri 6 ki te 12.

$$x=\frac{18}{6}$$

Whakawehe 18 ki te 6.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{6±12}{6} ina he tango te ±. Tango 12 mai i 6.

$$x=-\frac{6}{6}$$

Whakawehe -6 ki te 6.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=3 x=-1$$

Tē taea kia ōrite te tāupe x ki 3.

$$x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,3 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-3\right)\left(x+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-3,x+2.

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

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

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

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

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

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

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

Pahekotia te 2x me -5x, ka -3x.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x+2 ki te 3.

$$3x^{2}-3x-3=3x+6$$

Tangohia te 3x mai i ngā taha e rua.

$$3x^{2}-3x-3-3x=6$$

Pahekotia te -3x me -3x, ka -6x.

$$3x^{2}-6x-3=6$$

Me tāpiri te 3 ki ngā taha e rua.

$$3x^{2}-6x=6+3$$

Tāpirihia te 6 ki te 3, ka 9.

$$3x^{2}-6x=9$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}-6x}{3}=\frac{9}{3}$$

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

$$x^{2}+\left(-\frac{6}{3}\right)x=\frac{9}{3}$$

Whakawehe -6 ki te 3.

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

Whakawehe 9 ki te 3.

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

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

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

Tāpiri 3 ki te 1.

$$x^{2}-2x+1=4$$

Tauwehea te x^{2}-2x+1. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

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

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

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

Whakarūnātia.

$$x-1=2 x-1=-2$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$x=3 x=-1$$

Tē taea kia ōrite te tāupe x ki 3.

$$x=-1$$