Whakaoti i te 1/(x^2+3)+1/(x+4)=

$$\frac { 1 } { ( x ^ { 2 } + 3 ) } + \frac { 1 } { ( x + 4 ) } =$$

Aromātai

$$\frac{x^{2}+x+7}{\left(x+4\right)\left(x^{2}+3\right)}$$

Ngā Hipanga Rongoā Poto

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o x^{2}+3 me x+4 ko \left(x+4\right)\left(x^{2}+3\right). Whakareatia \frac{1}{x^{2}+3} ki te \frac{x+4}{x+4}. Whakareatia \frac{1}{x+4} ki te \frac{x^{2}+3}{x^{2}+3}.

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

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

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

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

$$\frac{x+7+x^{2}}{\left(x+4\right)\left(x^{2}+3\right)}$$

Whakarohaina te \left(x+4\right)\left(x^{2}+3\right).

$$\frac{x+7+x^{2}}{x^{3}+4x^{2}+3x+12}$$

Kimi Pārōnaki e ai ki x

$$-\frac{x^{4}+2x^{3}+22x^{2}+32x+9}{\left(\left(x+4\right)\left(x^{2}+3\right)\right)^{2}}$$