Whakaoti i te 1/x-3-2/x^2-9=

$$\frac { 1 } { x - 3 } - \frac { 2 } { x ^ { 2 } - 9 } =$$

Aromātai

$$\frac{x+1}{x^{2}-9}$$

Ngā Hipanga Rongoā Poto

Tauwehea te x^{2}-9.

$$\frac{1}{x-3}-\frac{2}{\left(x-3\right)\left(x+3\right)}$$

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

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

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

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

Whakakotahitia ngā kupu rite i x+3-2.

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

Whakarohaina te \left(x-3\right)\left(x+3\right).

$$\frac{x+1}{x^{2}-9}$$

Kimi Pārōnaki e ai ki x

$$\frac{-x^{2}-2x-9}{\left(x^{2}-9\right)^{2}}$$

Ngā Upane e Whakamahi ana i te Ture Pārōnaki mō te Otinga

Tauwehea te x^{2}-9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x-3}-\frac{2}{\left(x-3\right)\left(x+3\right)})$$

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

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

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

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

Whakakotahitia ngā kupu rite i x+3-2.

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

Whakaarohia te \left(x-3\right)\left(x+3\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+1}{x^{2}-9})$$

Mō ngā pānga e rua e taea ana te pārōnaki, ko te pārōnaki o te otinga o ngā pānga e rua ko te tauraro whakareatia ki te pārōnaki o te taurunga tango i te taurunga whakareatia ki te pārōnaki o te tauraro, ā, ka whakawehea te katoa ki te tauraro kua pūruatia.

$$\frac{\left(x^{2}-9\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{1}+1)-\left(x^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-9)}{\left(x^{2}-9\right)^{2}}$$

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}.

$$\frac{\left(x^{2}-9\right)x^{1-1}-\left(x^{1}+1\right)\times 2x^{2-1}}{\left(x^{2}-9\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{\left(x^{2}-9\right)x^{0}-\left(x^{1}+1\right)\times 2x^{1}}{\left(x^{2}-9\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{x^{2}x^{0}-9x^{0}-\left(x^{1}\times 2x^{1}+2x^{1}\right)}{\left(x^{2}-9\right)^{2}}$$

Hei whakarea pū o te pūtake ōrite, tāpiri ana taupū.

$$\frac{x^{2}-9x^{0}-\left(2x^{1+1}+2x^{1}\right)}{\left(x^{2}-9\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{x^{2}-9x^{0}-\left(2x^{2}+2x^{1}\right)}{\left(x^{2}-9\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{x^{2}-9x^{0}-2x^{2}-2x^{1}}{\left(x^{2}-9\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(1-2\right)x^{2}-9x^{0}-2x^{1}}{\left(x^{2}-9\right)^{2}}$$

Tango 2 mai i 1.

$$\frac{-x^{2}-9x^{0}-2x^{1}}{\left(x^{2}-9\right)^{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{-x^{2}-9x^{0}-2x}{\left(x^{2}-9\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{-x^{2}-9-2x}{\left(x^{2}-9\right)^{2}}$$