Whakaoti i te 1/x-3+2

$$\frac { 1 } { x - 3 } + 2$$

Aromātai

$$\frac{2x-5}{x-3}$$

Ngā Hipanga Rongoā Poto

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 2 ki te \frac{x-3}{x-3}.

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

Tā te mea he rite te tauraro o \frac{1}{x-3} me \frac{2\left(x-3\right)}{x-3}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

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

Mahia ngā whakarea i roto o 1+2\left(x-3\right).

$$\frac{1+2x-6}{x-3}$$

Whakakotahitia ngā kupu rite i 1+2x-6.

$$\frac{-5+2x}{x-3}$$

Kimi Pārōnaki e ai ki x

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

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 2 ki te \frac{x-3}{x-3}.

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

Tā te mea he rite te tauraro o \frac{1}{x-3} me \frac{2\left(x-3\right)}{x-3}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

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

Mahia ngā whakarea i roto o 1+2\left(x-3\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1+2x-6}{x-3})$$

Whakakotahitia ngā kupu rite i 1+2x-6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-5+2x}{x-3})$$

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.

$$\frac{\left(x^{1}-3\right)\frac{\mathrm{d}}{\mathrm{d}x}(2x^{1}-5)-\left(2x^{1}-5\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{1}-3)}{\left(x^{1}-3\right)^{2}}$$

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}.

$$\frac{\left(x^{1}-3\right)\times 2x^{1-1}-\left(2x^{1}-5\right)x^{1-1}}{\left(x^{1}-3\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{\left(x^{1}-3\right)\times 2x^{0}-\left(2x^{1}-5\right)x^{0}}{\left(x^{1}-3\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{x^{1}\times 2x^{0}-3\times 2x^{0}-\left(2x^{1}x^{0}-5x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

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

$$\frac{2x^{1}-3\times 2x^{0}-\left(2x^{1}-5x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{2x^{1}-6x^{0}-\left(2x^{1}-5x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{2x^{1}-6x^{0}-2x^{1}-\left(-5x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(2-2\right)x^{1}+\left(-6-\left(-5\right)\right)x^{0}}{\left(x^{1}-3\right)^{2}}$$

Tangohia te 2 i 2 me te -5 i te -6.

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

Mō tētahi kupu t, t^{1}=t.

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

Mō tētahi kupu t mahue te 0, t^{0}=1.

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