Whakaoti i te 6/1-4-3x/x^2-4=

$$\frac { 6 } { 1 - 4 } - \frac { 3 x } { x ^ { 2 } - 4 } =$$

Aromātai

$$\frac{8-3x-2x^{2}}{x^{2}-4}$$

Ngā Hipanga Rongoā Poto

Tangohia te 4 i te 1, ka -3.

$$\frac{6}{-3}-\frac{3x}{x^{2}-4}$$

Whakawehea te 6 ki te -3, kia riro ko -2.

$$-2-\frac{3x}{x^{2}-4}$$

Tauwehea te x^{2}-4.

$$-2-\frac{3x}{\left(x-2\right)\left(x+2\right)}$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia -2 ki te \frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}.

$$-\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3x}{\left(x-2\right)\left(x+2\right)}$$

Tā te mea he rite te tauraro o -\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} me \frac{3x}{\left(x-2\right)\left(x+2\right)}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{-2\left(x-2\right)\left(x+2\right)-3x}{\left(x-2\right)\left(x+2\right)}$$

Mahia ngā whakarea i roto o -2\left(x-2\right)\left(x+2\right)-3x.

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

Whakakotahitia ngā kupu rite i -2x^{2}-4x+4x+8-3x.

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

Whakarohaina te \left(x-2\right)\left(x+2\right).

$$\frac{-2x^{2}-3x+8}{x^{2}-4}$$

Tauwehe

$$-\frac{2\left(x-\frac{-\sqrt{73}-3}{4}\right)\left(x-\frac{\sqrt{73}-3}{4}\right)}{x^{2}-4}$$