Whakaoti i te x^2/x^2+6x+9-4/x^2-9=

$$\frac { x ^ { 2 } } { x ^ { 2 } + 6 x + 9 } - \frac { 4 } { x ^ { 2 } - 9 } =$$

Aromātai

$$\frac{x^{3}-3x^{2}-4x-12}{\left(x-3\right)\left(x+3\right)^{2}}$$

Ngā Hipanga Rongoā Poto

Tauwehea te x^{2}+6x+9. Tauwehea te x^{2}-9.

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o \left(x+3\right)^{2} me \left(x-3\right)\left(x+3\right) ko \left(x-3\right)\left(x+3\right)^{2}. Whakareatia \frac{x^{2}}{\left(x+3\right)^{2}} ki te \frac{x-3}{x-3}. Whakareatia \frac{4}{\left(x-3\right)\left(x+3\right)} ki te \frac{x+3}{x+3}.

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

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

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

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

$$\frac{x^{3}-3x^{2}-4x-12}{\left(x-3\right)\left(x+3\right)^{2}}$$

Whakarohaina te \left(x-3\right)\left(x+3\right)^{2}.

$$\frac{x^{3}-3x^{2}-4x-12}{x^{3}+3x^{2}-9x-27}$$

Kimi Pārōnaki e ai ki x

$$\frac{2x\left(3x^{2}-14x+39\right)}{\left(x-3\right)^{2}\left(x+3\right)^{3}}$$