Whakaoti i te 6/a^2-9+1/a+3

$$\frac { 6 } { a ^ { 2 } - 9 } + \frac { 1 } { a + 3 }$$

Aromātai

$$\frac{1}{a-3}$$

Ngā Hipanga Rongoā Poto

Tauwehea te a^{2}-9.

$$\frac{6}{\left(a-3\right)\left(a+3\right)}+\frac{1}{a+3}$$

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

$$\frac{6}{\left(a-3\right)\left(a+3\right)}+\frac{a-3}{\left(a-3\right)\left(a+3\right)}$$

Tā te mea he rite te tauraro o \frac{6}{\left(a-3\right)\left(a+3\right)} me \frac{a-3}{\left(a-3\right)\left(a+3\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{6+a-3}{\left(a-3\right)\left(a+3\right)}$$

Whakakotahitia ngā kupu rite i 6+a-3.

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

Me whakakore tahi te a+3 i te taurunga me te tauraro.

$$\frac{1}{a-3}$$

Kimi Pārōnaki e ai ki a

$$-\frac{1}{\left(a-3\right)^{2}}$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Tauwehea te a^{2}-9.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{6}{\left(a-3\right)\left(a+3\right)}+\frac{1}{a+3})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{6}{\left(a-3\right)\left(a+3\right)}+\frac{a-3}{\left(a-3\right)\left(a+3\right)})$$

Tā te mea he rite te tauraro o \frac{6}{\left(a-3\right)\left(a+3\right)} me \frac{a-3}{\left(a-3\right)\left(a+3\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{6+a-3}{\left(a-3\right)\left(a+3\right)})$$

Whakakotahitia ngā kupu rite i 6+a-3.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{3+a}{\left(a-3\right)\left(a+3\right)})$$

Me whakakore tahi te a+3 i te taurunga me te tauraro.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{a-3})$$

Mēnā ko F te hanganga o ngā pānga e rua e taea ana te pārōnaki f\left(u\right) me u=g\left(x\right), arā, mēnā ko F\left(x\right)=f\left(g\left(x\right)\right), ko te pārōnaki o F te pārōnaki o f e ai ki u whakareatia te pārōnaki o g e ai ki x, arā, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$-\left(a^{1}-3\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}a}(a^{1}-3)$$

Ko te pārōnaki o tētahi pūrau ko te tapeke o ngā pārōnaki o ōna kīanga tau. Ko te pārōnaki o tētahi kīanga tau pūmau ko 0. Ko te pārōnaki o te ax^{n} ko te nax^{n-1}.

$$-\left(a^{1}-3\right)^{-2}a^{1-1}$$

Whakarūnātia.

$$-a^{0}\left(a^{1}-3\right)^{-2}$$

Mō tētahi kupu t, t^{1}=t.

$$-a^{0}\left(a-3\right)^{-2}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$-\left(a-3\right)^{-2}$$