Whakaoti i te 12/4+x+12/4-x=8

$$\frac { 12 } { 4 + x } + \frac { 12 } { 4 - x } = 8$$

Whakaoti mō x

$$x=-2$$

$$x=2$$

Ngā Upane mā te Kimi Pūtakerua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -4,4 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-4\right)\left(x+4\right), arā, te tauraro pātahi he tino iti rawa te kitea o 4+x,4-x.

$$\left(x-4\right)\times 12-\left(4+x\right)\times 12=8\left(x-4\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-4 ki te 12.

$$12x-48-\left(4+x\right)\times 12=8\left(x-4\right)\left(x+4\right)$$

Whakareatia te -1 ki te 12, ka -12.

$$12x-48-12\left(4+x\right)=8\left(x-4\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -12 ki te 4+x.

$$12x-48-48-12x=8\left(x-4\right)\left(x+4\right)$$

Tangohia te 48 i te -48, ka -96.

$$12x-96-12x=8\left(x-4\right)\left(x+4\right)$$

Pahekotia te 12x me -12x, ka 0.

$$-96=8\left(x-4\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 8 ki te x-4.

$$-96=\left(8x-32\right)\left(x+4\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 8x-32 ki te x+4 ka whakakotahi i ngā kupu rite.

$$-96=8x^{2}-128$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$8x^{2}-128=-96$$

Me tāpiri te 128 ki ngā taha e rua.

$$8x^{2}=-96+128$$

Tāpirihia te -96 ki te 128, ka 32.

$$8x^{2}=32$$

Whakawehea ngā taha e rua ki te 8.

$$x^{2}=\frac{32}{8}$$

Whakawehea te 32 ki te 8, kia riro ko 4.

$$x^{2}=4$$

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

$$x=2 x=-2$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -4,4 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(x-4\right)\left(x+4\right), arā, te tauraro pātahi he tino iti rawa te kitea o 4+x,4-x.

$$\left(x-4\right)\times 12-\left(4+x\right)\times 12=8\left(x-4\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te x-4 ki te 12.

$$12x-48-\left(4+x\right)\times 12=8\left(x-4\right)\left(x+4\right)$$

Whakareatia te -1 ki te 12, ka -12.

$$12x-48-12\left(4+x\right)=8\left(x-4\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -12 ki te 4+x.

$$12x-48-48-12x=8\left(x-4\right)\left(x+4\right)$$

Tangohia te 48 i te -48, ka -96.

$$12x-96-12x=8\left(x-4\right)\left(x+4\right)$$

Pahekotia te 12x me -12x, ka 0.

$$-96=8\left(x-4\right)\left(x+4\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 8 ki te x-4.

$$-96=\left(8x-32\right)\left(x+4\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 8x-32 ki te x+4 ka whakakotahi i ngā kupu rite.

$$-96=8x^{2}-128$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$8x^{2}-128=-96$$

Me tāpiri te 96 ki ngā taha e rua.

$$8x^{2}-128+96=0$$

Tāpirihia te -128 ki te 96, ka -32.

$$8x^{2}-32=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, 0 mō b, me -32 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 8\left(-32\right)}}{2\times 8}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\times 8\left(-32\right)}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{0±\sqrt{-32\left(-32\right)}}{2\times 8}$$

Whakareatia -32 ki te -32.

$$x=\frac{0±\sqrt{1024}}{2\times 8}$$

Tuhia te pūtakerua o te 1024.

$$x=\frac{0±32}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{0±32}{16}$$

Nā, me whakaoti te whārite x=\frac{0±32}{16} ina he tāpiri te ±. Whakawehe 32 ki te 16.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{0±32}{16} ina he tango te ±. Whakawehe -32 ki te 16.

$$x=-2$$

Kua oti te whārite te whakatau.

$$x=2 x=-2$$