Whakaoti i te (x+3)/(x-2)-8/3=(x+2)/(x-1)

$$\frac { ( x + 3 ) } { ( x - 2 ) } - \frac { 8 } { 3 } = \frac { ( x + 2 ) } { ( x - 1 ) }$$

Whakaoti mō x

$$x = \frac{13}{4} = 3\frac{1}{4} = 3.25$$

$$x=\frac{1}{2}=0.5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 1,2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 3\left(x-2\right)\left(x-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-2,3,x-1.

$$\left(3x-3\right)\left(x+3\right)+3\left(x-2\right)\left(x-1\right)\left(-\frac{8}{3}\right)=\left(3x-6\right)\left(x+2\right)$$

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

$$3x^{2}+6x-9+3\left(x-2\right)\left(x-1\right)\left(-\frac{8}{3}\right)=\left(3x-6\right)\left(x+2\right)$$

Whakareatia te 3 ki te -\frac{8}{3}, ka -8.

$$3x^{2}+6x-9-8\left(x-2\right)\left(x-1\right)=\left(3x-6\right)\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -8 ki te x-2.

$$3x^{2}+6x-9+\left(-8x+16\right)\left(x-1\right)=\left(3x-6\right)\left(x+2\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -8x+16 ki te x-1 ka whakakotahi i ngā kupu rite.

$$3x^{2}+6x-9-8x^{2}+24x-16=\left(3x-6\right)\left(x+2\right)$$

Pahekotia te 3x^{2} me -8x^{2}, ka -5x^{2}.

$$-5x^{2}+6x-9+24x-16=\left(3x-6\right)\left(x+2\right)$$

Pahekotia te 6x me 24x, ka 30x.

$$-5x^{2}+30x-9-16=\left(3x-6\right)\left(x+2\right)$$

Tangohia te 16 i te -9, ka -25.

$$-5x^{2}+30x-25=\left(3x-6\right)\left(x+2\right)$$

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

$$-5x^{2}+30x-25=3x^{2}-12$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$-5x^{2}+30x-25-3x^{2}=-12$$

Pahekotia te -5x^{2} me -3x^{2}, ka -8x^{2}.

$$-8x^{2}+30x-25=-12$$

Me tāpiri te 12 ki ngā taha e rua.

$$-8x^{2}+30x-25+12=0$$

Tāpirihia te -25 ki te 12, ka -13.

$$-8x^{2}+30x-13=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -8 mō a, 30 mō b, me -13 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-8\right)\left(-13\right)}}{2\left(-8\right)}$$

Pūrua 30.

$$x=\frac{-30±\sqrt{900-4\left(-8\right)\left(-13\right)}}{2\left(-8\right)}$$

Whakareatia -4 ki te -8.

$$x=\frac{-30±\sqrt{900+32\left(-13\right)}}{2\left(-8\right)}$$

Whakareatia 32 ki te -13.

$$x=\frac{-30±\sqrt{900-416}}{2\left(-8\right)}$$

Tāpiri 900 ki te -416.

$$x=\frac{-30±\sqrt{484}}{2\left(-8\right)}$$

Tuhia te pūtakerua o te 484.

$$x=\frac{-30±22}{2\left(-8\right)}$$

Whakareatia 2 ki te -8.

$$x=\frac{-30±22}{-16}$$

Nā, me whakaoti te whārite x=\frac{-30±22}{-16} ina he tāpiri te ±. Tāpiri -30 ki te 22.

$$x=-\frac{8}{-16}$$

Whakahekea te hautanga \frac{-8}{-16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$x=\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-30±22}{-16} ina he tango te ±. Tango 22 mai i -30.

$$x=-\frac{52}{-16}$$

Whakahekea te hautanga \frac{-52}{-16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{13}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{2} x=\frac{13}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 1,2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 3\left(x-2\right)\left(x-1\right), arā, te tauraro pātahi he tino iti rawa te kitea o x-2,3,x-1.

$$\left(3x-3\right)\left(x+3\right)+3\left(x-2\right)\left(x-1\right)\left(-\frac{8}{3}\right)=\left(3x-6\right)\left(x+2\right)$$

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

$$3x^{2}+6x-9+3\left(x-2\right)\left(x-1\right)\left(-\frac{8}{3}\right)=\left(3x-6\right)\left(x+2\right)$$

Whakareatia te 3 ki te -\frac{8}{3}, ka -8.

$$3x^{2}+6x-9-8\left(x-2\right)\left(x-1\right)=\left(3x-6\right)\left(x+2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -8 ki te x-2.

$$3x^{2}+6x-9+\left(-8x+16\right)\left(x-1\right)=\left(3x-6\right)\left(x+2\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -8x+16 ki te x-1 ka whakakotahi i ngā kupu rite.

$$3x^{2}+6x-9-8x^{2}+24x-16=\left(3x-6\right)\left(x+2\right)$$

Pahekotia te 3x^{2} me -8x^{2}, ka -5x^{2}.

$$-5x^{2}+6x-9+24x-16=\left(3x-6\right)\left(x+2\right)$$

Pahekotia te 6x me 24x, ka 30x.

$$-5x^{2}+30x-9-16=\left(3x-6\right)\left(x+2\right)$$

Tangohia te 16 i te -9, ka -25.

$$-5x^{2}+30x-25=\left(3x-6\right)\left(x+2\right)$$

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

$$-5x^{2}+30x-25=3x^{2}-12$$

Tangohia te 3x^{2} mai i ngā taha e rua.

$$-5x^{2}+30x-25-3x^{2}=-12$$

Pahekotia te -5x^{2} me -3x^{2}, ka -8x^{2}.

$$-8x^{2}+30x-25=-12$$

Me tāpiri te 25 ki ngā taha e rua.

$$-8x^{2}+30x=-12+25$$

Tāpirihia te -12 ki te 25, ka 13.

$$-8x^{2}+30x=13$$

Whakawehea ngā taha e rua ki te -8.

$$\frac{-8x^{2}+30x}{-8}=\frac{13}{-8}$$

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

$$x^{2}+\frac{30}{-8}x=\frac{13}{-8}$$

Whakahekea te hautanga \frac{30}{-8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}-\frac{15}{4}x=\frac{13}{-8}$$

Whakawehe 13 ki te -8.

$$x^{2}-\frac{15}{4}x=-\frac{13}{8}$$

Whakawehea te -\frac{15}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{15}{8}. Nā, tāpiria te pūrua o te -\frac{15}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{15}{4}x+\left(-\frac{15}{8}\right)^{2}=-\frac{13}{8}+\left(-\frac{15}{8}\right)^{2}$$

Pūruatia -\frac{15}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{15}{4}x+\frac{225}{64}=-\frac{13}{8}+\frac{225}{64}$$

Tāpiri -\frac{13}{8} ki te \frac{225}{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.

$$x^{2}-\frac{15}{4}x+\frac{225}{64}=\frac{121}{64}$$

Tauwehea x^{2}-\frac{15}{4}x+\frac{225}{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}.

$$\left(x-\frac{15}{8}\right)^{2}=\frac{121}{64}$$

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

$$\sqrt{\left(x-\frac{15}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Whakarūnātia.

$$x-\frac{15}{8}=\frac{11}{8} x-\frac{15}{8}=-\frac{11}{8}$$

Me tāpiri \frac{15}{8} ki ngā taha e rua o te whārite.

$$x=\frac{13}{4} x=\frac{1}{2}$$