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

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

Whakaoti mō x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2+x,2-x.

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

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

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

Tuhia te \left(2+x\right)\times \frac{x}{2} hei hautanga kotahi.

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

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

$$x^{2}-2x=-\frac{2x+x^{2}}{2}$$

Whakawehea ia wā o 2x+x^{2} ki te 2, kia riro ko x+\frac{1}{2}x^{2}.

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

Hei kimi i te tauaro o x+\frac{1}{2}x^{2}, kimihia te tauaro o ia taurangi.

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

Me tāpiri te x ki ngā taha e rua.

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

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

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

Me tāpiri te \frac{1}{2}x^{2} ki ngā taha e rua.

$$x^{2}-x+\frac{1}{2}x^{2}=0$$

Pahekotia te x^{2} me \frac{1}{2}x^{2}, ka \frac{3}{2}x^{2}.

$$\frac{3}{2}x^{2}-x=0$$

Tauwehea te x.

$$x\left(\frac{3}{2}x-1\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te \frac{3x}{2}-1=0.

$$x=0 x=\frac{2}{3}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2+x,2-x.

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

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

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

Tuhia te \left(2+x\right)\times \frac{x}{2} hei hautanga kotahi.

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

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

$$x^{2}-2x=-\frac{2x+x^{2}}{2}$$

Whakawehea ia wā o 2x+x^{2} ki te 2, kia riro ko x+\frac{1}{2}x^{2}.

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

Hei kimi i te tauaro o x+\frac{1}{2}x^{2}, kimihia te tauaro o ia taurangi.

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

Me tāpiri te x ki ngā taha e rua.

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

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

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

Me tāpiri te \frac{1}{2}x^{2} ki ngā taha e rua.

$$x^{2}-x+\frac{1}{2}x^{2}=0$$

Pahekotia te x^{2} me \frac{1}{2}x^{2}, ka \frac{3}{2}x^{2}.

$$\frac{3}{2}x^{2}-x=0$$

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

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

Tuhia te pūtakerua o te 1.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2\times \frac{3}{2}}$$

Whakareatia 2 ki te \frac{3}{2}.

$$x=\frac{1±1}{3}$$

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

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

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

$$x=\frac{0}{3}$$

Whakawehe 0 ki te 3.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{2}{3} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -2,2 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+2\right), arā, te tauraro pātahi he tino iti rawa te kitea o 2+x,2-x.

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

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

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

Tuhia te \left(2+x\right)\times \frac{x}{2} hei hautanga kotahi.

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

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

$$x^{2}-2x=-\frac{2x+x^{2}}{2}$$

Whakawehea ia wā o 2x+x^{2} ki te 2, kia riro ko x+\frac{1}{2}x^{2}.

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

Hei kimi i te tauaro o x+\frac{1}{2}x^{2}, kimihia te tauaro o ia taurangi.

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

Me tāpiri te x ki ngā taha e rua.

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

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

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

Me tāpiri te \frac{1}{2}x^{2} ki ngā taha e rua.

$$x^{2}-x+\frac{1}{2}x^{2}=0$$

Pahekotia te x^{2} me \frac{1}{2}x^{2}, ka \frac{3}{2}x^{2}.

$$\frac{3}{2}x^{2}-x=0$$

Whakawehea ngā taha e rua o te whārite ki te \frac{3}{2}, he ōrite ki te whakarea i ngā taha e rua ki te tau huripoki o te hautanga.

$$\frac{\frac{3}{2}x^{2}-x}{\frac{3}{2}}=\frac{0}{\frac{3}{2}}$$

Mā te whakawehe ki te \frac{3}{2} ka wetekia te whakareanga ki te \frac{3}{2}.

$$x^{2}+\left(-\frac{1}{\frac{3}{2}}\right)x=\frac{0}{\frac{3}{2}}$$

Whakawehe -1 ki te \frac{3}{2} mā te whakarea -1 ki te tau huripoki o \frac{3}{2}.

$$x^{2}-\frac{2}{3}x=\frac{0}{\frac{3}{2}}$$

Whakawehe 0 ki te \frac{3}{2} mā te whakarea 0 ki te tau huripoki o \frac{3}{2}.

$$x^{2}-\frac{2}{3}x=0$$

Whakawehea te -\frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{3}. Nā, tāpiria te pūrua o te -\frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tauwehea x^{2}-\frac{2}{3}x+\frac{1}{9}. 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}.

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

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

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

Whakarūnātia.

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

Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

$$x=\frac{2}{3} x=0$$