Whakaoti i te 3x-7/x+5=x-3/x+2

$$\frac { 3 x - 7 } { x + 5 } = \frac { x - 3 } { x + 2 }$$

Whakaoti mō x

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

$$x=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -5,-2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x+2\right)\left(x+5\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+5,x+2.

$$\left(x+2\right)\left(3x-7\right)=\left(x+5\right)\left(x-3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te 3x-7 ka whakakotahi i ngā kupu rite.

$$3x^{2}-x-14=\left(x+5\right)\left(x-3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+5 ki te x-3 ka whakakotahi i ngā kupu rite.

$$3x^{2}-x-14=x^{2}+2x-15$$

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

$$3x^{2}-x-14-x^{2}=2x-15$$

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

$$2x^{2}-x-14=2x-15$$

Tangohia te 2x mai i ngā taha e rua.

$$2x^{2}-x-14-2x=-15$$

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

$$2x^{2}-3x-14=-15$$

Me tāpiri te 15 ki ngā taha e rua.

$$2x^{2}-3x-14+15=0$$

Tāpirihia te -14 ki te 15, ka 1.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -3 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -3.

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

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\times 2}$$

Tāpiri 9 ki te -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2\times 2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-3\right)±1}{2\times 2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±1}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{3±1}{4}$$

Nā, me whakaoti te whārite x=\frac{3±1}{4} ina he tāpiri te ±. Tāpiri 3 ki te 1.

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

Whakawehe 4 ki te 4.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{3±1}{4} ina he tango te ±. Tango 1 mai i 3.

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

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -5,-2 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x+2\right)\left(x+5\right), arā, te tauraro pātahi he tino iti rawa te kitea o x+5,x+2.

$$\left(x+2\right)\left(3x-7\right)=\left(x+5\right)\left(x-3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+2 ki te 3x-7 ka whakakotahi i ngā kupu rite.

$$3x^{2}-x-14=\left(x+5\right)\left(x-3\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te x+5 ki te x-3 ka whakakotahi i ngā kupu rite.

$$3x^{2}-x-14=x^{2}+2x-15$$

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

$$3x^{2}-x-14-x^{2}=2x-15$$

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

$$2x^{2}-x-14=2x-15$$

Tangohia te 2x mai i ngā taha e rua.

$$2x^{2}-x-14-2x=-15$$

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

$$2x^{2}-3x-14=-15$$

Me tāpiri te 14 ki ngā taha e rua.

$$2x^{2}-3x=-15+14$$

Tāpirihia te -15 ki te 14, ka -1.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehea te -\frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{4}. Nā, tāpiria te pūrua o te -\frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Tāpiri -\frac{1}{2} ki te \frac{9}{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.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Tauwehea te x^{2}-\frac{3}{2}x+\frac{9}{16}. 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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

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

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Whakarūnātia.

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

Me tāpiri \frac{3}{4} ki ngā taha e rua o te whārite.

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