Whakaoti i te fff:x^4-2x^2-8=

$$f f f : x ^ { 4 } - 2 x ^ { 2 } - 8 =$$

Aromātai

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

Ngā Hipanga Rongoā Poto

Whakareatia te f ki te f, ka f^{2}.

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

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 2 me te 1 kia riro ai te 3.

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia -2x^{2}-8 ki te \frac{x^{4}}{x^{4}}.

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

Tā te mea he rite te tauraro o \frac{f^{3}}{x^{4}} me \frac{\left(-2x^{2}-8\right)x^{4}}{x^{4}}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

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

Mahia ngā whakarea i roto o f^{3}+\left(-2x^{2}-8\right)x^{4}.

$$\frac{f^{3}-2x^{6}-8x^{4}}{x^{4}}$$

Kimi Pārōnaki e ai ki x

$$-4x-\frac{4f^{3}}{x^{5}}$$