Whakaoti i te 9/2x+1-8x/2x-1=-4

$$\frac { 9 } { 2 x + 1 } - \frac { 8 x } { 2 x - 1 } = - 4$$

Whakaoti mō x

$$x = \frac{13}{10} = 1\frac{3}{10} = 1.3$$

Ngā Upane mō te Whakaoti Whārite Rārangi

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

$$\left(2x-1\right)\times 9-\left(2x+1\right)\times 8x=-4\left(2x-1\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x-1 ki te 9.

$$18x-9-\left(2x+1\right)\times 8x=-4\left(2x-1\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x+1 ki te 8.

$$18x-9-\left(16x+8\right)x=-4\left(2x-1\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 16x+8 ki te x.

$$18x-9-\left(16x^{2}+8x\right)=-4\left(2x-1\right)\left(2x+1\right)$$

Hei kimi i te tauaro o 16x^{2}+8x, kimihia te tauaro o ia taurangi.

$$18x-9-16x^{2}-8x=-4\left(2x-1\right)\left(2x+1\right)$$

Pahekotia te 18x me -8x, ka 10x.

$$10x-9-16x^{2}=-4\left(2x-1\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te -4 ki te 2x-1.

$$10x-9-16x^{2}=\left(-8x+4\right)\left(2x+1\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te -8x+4 ki te 2x+1 ka whakakotahi i ngā kupu rite.

$$10x-9-16x^{2}=-16x^{2}+4$$

Me tāpiri te 16x^{2} ki ngā taha e rua.

$$10x-9-16x^{2}+16x^{2}=4$$

Pahekotia te -16x^{2} me 16x^{2}, ka 0.

$$10x-9=4$$

Me tāpiri te 9 ki ngā taha e rua.

$$10x=4+9$$

Tāpirihia te 4 ki te 9, ka 13.

$$10x=13$$

Whakawehea ngā taha e rua ki te 10.

$$x=\frac{13}{10}$$