Whakaoti i te (2x-1)(2x+1)/3=3x-2/6+x^2/3

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

Whakaoti mō x

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

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Me whakarea ngā taha e rua o te whārite ki te 6, arā, te tauraro pātahi he tino iti rawa te kitea o 3,6.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 2x-1.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x-2 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$8x^{2}-2=3x-2+2x^{2}$$

Tangohia te 3x mai i ngā taha e rua.

$$8x^{2}-2-3x=-2+2x^{2}$$

Tangohia te -2 mai i ngā taha e rua.

$$8x^{2}-2-3x-\left(-2\right)=2x^{2}$$

Ko te tauaro o -2 ko 2.

$$8x^{2}-2-3x+2=2x^{2}$$

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

$$8x^{2}-2-3x+2-2x^{2}=0$$

Tāpirihia te -2 ki te 2, ka 0.

$$8x^{2}-3x-2x^{2}=0$$

Pahekotia te 8x^{2} me -2x^{2}, ka 6x^{2}.

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

Tauwehea te x.

$$x\left(6x-3\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 6x-3=0.

$$x=0 x=\frac{1}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua o te whārite ki te 6, arā, te tauraro pātahi he tino iti rawa te kitea o 3,6.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 2x-1.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x-2 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$8x^{2}-2=3x-2+2x^{2}$$

Tangohia te 3x mai i ngā taha e rua.

$$8x^{2}-2-3x=-2+2x^{2}$$

Tangohia te -2 mai i ngā taha e rua.

$$8x^{2}-2-3x-\left(-2\right)=2x^{2}$$

Ko te tauaro o -2 ko 2.

$$8x^{2}-2-3x+2=2x^{2}$$

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

$$8x^{2}-2-3x+2-2x^{2}=0$$

Tāpirihia te -2 ki te 2, ka 0.

$$8x^{2}-3x-2x^{2}=0$$

Pahekotia te 8x^{2} me -2x^{2}, ka 6x^{2}.

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

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

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

Tuhia te pūtakerua o te \left(-3\right)^{2}.

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

Ko te tauaro o -3 ko 3.

$$x=\frac{3±3}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{3±3}{12}$$

Nā, me whakaoti te whārite x=\frac{3±3}{12} ina he tāpiri te ±. Tāpiri 3 ki te 3.

$$x=\frac{6}{12}$$

Whakahekea te hautanga \frac{6}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Nā, me whakaoti te whārite x=\frac{3±3}{12} ina he tango te ±. Tango 3 mai i 3.

$$x=\frac{0}{12}$$

Whakawehe 0 ki te 12.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{2} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Me whakarea ngā taha e rua o te whārite ki te 6, arā, te tauraro pātahi he tino iti rawa te kitea o 3,6.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te 2x-1.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x-2 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$8x^{2}-2=3x-2+2x^{2}$$

Tangohia te 3x mai i ngā taha e rua.

$$8x^{2}-2-3x=-2+2x^{2}$$

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

$$8x^{2}-2-3x-2x^{2}=-2$$

Pahekotia te 8x^{2} me -2x^{2}, ka 6x^{2}.

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

Me tāpiri te 2 ki ngā taha e rua.

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

Tāpirihia te -2 ki te 2, ka 0.

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

Whakawehea ngā taha e rua ki te 6.

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

Mā te whakawehe ki te 6 ka wetekia te whakareanga ki te 6.

$$x^{2}+\left(-\frac{3}{6}\right)x=\frac{0}{6}$$

Whakahekea te hautanga \frac{-3}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$x^{2}-\frac{1}{2}x=\frac{0}{6}$$

Whakawehe 0 ki te 6.

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

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}.

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

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

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

Whakarūnātia.

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

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{1}{2} x=0$$