Whakaoti i te x^2+2/3-x^2+1/4=x+5/12

$$\frac { x ^ { 2 } + 2 } { 3 } - \frac { x ^ { 2 } + 1 } { 4 } = \frac { x + 5 } { 12 }$$

Whakaoti mō x

$$x=1$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Me whakarea ngā taha e rua o te whārite ki te 12, arā, te tauraro pātahi he tino iti rawa te kitea o 3,4,12.

$$4\left(x^{2}+2\right)-3\left(x^{2}+1\right)=x+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+2.

$$4x^{2}+8-3\left(x^{2}+1\right)=x+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x^{2}+1.

$$4x^{2}+8-3x^{2}-3=x+5$$

Pahekotia te 4x^{2} me -3x^{2}, ka x^{2}.

$$x^{2}+8-3=x+5$$

Tangohia te 3 i te 8, ka 5.

$$x^{2}+5=x+5$$

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 5 mai i ngā taha e rua.

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

Tangohia te 5 i te 5, ka 0.

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

Tauwehea te x.

$$x\left(x-1\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te x-1=0.

$$x=0 x=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua o te whārite ki te 12, arā, te tauraro pātahi he tino iti rawa te kitea o 3,4,12.

$$4\left(x^{2}+2\right)-3\left(x^{2}+1\right)=x+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+2.

$$4x^{2}+8-3\left(x^{2}+1\right)=x+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x^{2}+1.

$$4x^{2}+8-3x^{2}-3=x+5$$

Pahekotia te 4x^{2} me -3x^{2}, ka x^{2}.

$$x^{2}+8-3=x+5$$

Tangohia te 3 i te 8, ka 5.

$$x^{2}+5=x+5$$

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 5 mai i ngā taha e rua.

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

Tangohia te 5 i te 5, ka 0.

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

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

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

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-1\right)±1}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2}$$

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

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

Whakawehe 2 ki te 2.

$$x=1$$

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

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

Whakawehe 0 ki te 2.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=1 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Me whakarea ngā taha e rua o te whārite ki te 12, arā, te tauraro pātahi he tino iti rawa te kitea o 3,4,12.

$$4\left(x^{2}+2\right)-3\left(x^{2}+1\right)=x+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x^{2}+2.

$$4x^{2}+8-3\left(x^{2}+1\right)=x+5$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te x^{2}+1.

$$4x^{2}+8-3x^{2}-3=x+5$$

Pahekotia te 4x^{2} me -3x^{2}, ka x^{2}.

$$x^{2}+8-3=x+5$$

Tangohia te 3 i te 8, ka 5.

$$x^{2}+5=x+5$$

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 5 mai i ngā taha e rua.

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

Tangohia te 5 i te 5, ka 0.

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

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

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

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

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

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

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

Whakarūnātia.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=1 x=0$$