Whakaoti i te 1+2x/1+3x-1-2x/1-3x=-3x+14/1-9x^2

$$\frac { 1 + 2 x } { 1 + 3 x } - \frac { 1 - 2 x } { 1 - 3 x } = - \frac { 3 x + 14 } { 1 - 9 x ^ { 2 } }$$

Whakaoti mō x

$$x=-14$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tē taea kia ōrite te tāupe x ki tētahi o ngā uara -\frac{1}{3},\frac{1}{3} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te \left(3x-1\right)\left(3x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 1+3x,1-3x,1-9x^{2}.

$$\left(3x-1\right)\left(1+2x\right)-\left(-1-3x\right)\left(1-2x\right)=3x+14$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x-1 ki te 1+2x ka whakakotahi i ngā kupu rite.

$$x+6x^{2}-1-\left(-1-3x\right)\left(1-2x\right)=3x+14$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -1-3x ki te 1-2x ka whakakotahi i ngā kupu rite.

$$x+6x^{2}-1-\left(-1-x+6x^{2}\right)=3x+14$$

Hei kimi i te tauaro o -1-x+6x^{2}, kimihia te tauaro o ia taurangi.

$$x+6x^{2}-1+1+x-6x^{2}=3x+14$$

Tāpirihia te -1 ki te 1, ka 0.

$$x+6x^{2}+x-6x^{2}=3x+14$$

Pahekotia te x me x, ka 2x.

$$2x+6x^{2}-6x^{2}=3x+14$$

Pahekotia te 6x^{2} me -6x^{2}, ka 0.

$$2x=3x+14$$

Tangohia te 3x mai i ngā taha e rua.

$$2x-3x=14$$

Pahekotia te 2x me -3x, ka -x.

$$-x=14$$

Me whakarea ngā taha e rua ki te -1.

$$x=-14$$