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

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

Whakaoti mō x

$$x=2$$

$$x=3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,\frac{3}{2} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(2x-3\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x-3,x.

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

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

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

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

$$x^{2}-x=4x-6$$

Tangohia te 4x mai i ngā taha e rua.

$$x^{2}-x-4x=-6$$

Pahekotia te -x me -4x, ka -5x.

$$x^{2}-5x=-6$$

Me tāpiri te 6 ki ngā taha e rua.

$$x^{2}-5x+6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Tāpiri 25 ki te -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-5\right)±1}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±1}{2}$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tāpiri te ±. Tāpiri 5 ki te 1.

$$x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tango te ±. Tango 1 mai i 5.

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

Whakawehe 4 ki te 2.

$$x=2$$

Kua oti te whārite te whakatau.

$$x=3 x=2$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara 0,\frac{3}{2} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te x\left(2x-3\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2x-3,x.

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

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

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

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

$$x^{2}-x=4x-6$$

Tangohia te 4x mai i ngā taha e rua.

$$x^{2}-x-4x=-6$$

Pahekotia te -x me -4x, ka -5x.

$$x^{2}-5x=-6$$

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Tāpiri -6 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Tauwehea x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=3 x=2$$