Whakaoti i te 5/x+1+6/x-1

$$\frac{ 5 }{ x+1 } + \frac{ 6 }{ x-1 }$$

Aromātai

$$\frac{11x+1}{x^{2}-1}$$

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+1 me x-1 ko \left(x-1\right)\left(x+1\right). Whakareatia \frac{5}{x+1} ki te \frac{x-1}{x-1}. Whakareatia \frac{6}{x-1} ki te \frac{x+1}{x+1}.

$$\frac{5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{6\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}$$

Tā te mea he rite te tauraro o \frac{5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} me \frac{6\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{5\left(x-1\right)+6\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}$$

Mahia ngā whakarea i roto o 5\left(x-1\right)+6\left(x+1\right).

$$\frac{5x-5+6x+6}{\left(x-1\right)\left(x+1\right)}$$

Whakakotahitia ngā kupu rite i 5x-5+6x+6.

$$\frac{11x+1}{\left(x-1\right)\left(x+1\right)}$$

Whakarohaina te \left(x-1\right)\left(x+1\right).

$$\frac{11x+1}{x^{2}-1}$$

Kimi Pārōnaki e ai ki x

$$-\frac{11x^{2}+2x+11}{\left(x^{2}-1\right)^{2}}$$

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. Ko te taurea pātahi iti rawa o x+1 me x-1 ko \left(x-1\right)\left(x+1\right). Whakareatia \frac{5}{x+1} ki te \frac{x-1}{x-1}. Whakareatia \frac{6}{x-1} ki te \frac{x+1}{x+1}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{6\left(x+1\right)}{\left(x-1\right)\left(x+1\right)})$$

Tā te mea he rite te tauraro o \frac{5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} me \frac{6\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{5\left(x-1\right)+6\left(x+1\right)}{\left(x-1\right)\left(x+1\right)})$$

Mahia ngā whakarea i roto o 5\left(x-1\right)+6\left(x+1\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{5x-5+6x+6}{\left(x-1\right)\left(x+1\right)})$$

Whakakotahitia ngā kupu rite i 5x-5+6x+6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{11x+1}{\left(x-1\right)\left(x+1\right)})$$

Whakaarohia te \left(x-1\right)\left(x+1\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}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{11x+1}{x^{2}-1^{2}})$$

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{11x+1}{x^{2}-1})$$

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.

$$\frac{\left(x^{2}-1\right)\frac{\mathrm{d}}{\mathrm{d}x}(11x^{1}+1)-\left(11x^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-1)}{\left(x^{2}-1\right)^{2}}$$

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}.

$$\frac{\left(x^{2}-1\right)\times 11x^{1-1}-\left(11x^{1}+1\right)\times 2x^{2-1}}{\left(x^{2}-1\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{\left(x^{2}-1\right)\times 11x^{0}-\left(11x^{1}+1\right)\times 2x^{1}}{\left(x^{2}-1\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{x^{2}\times 11x^{0}-11x^{0}-\left(11x^{1}\times 2x^{1}+2x^{1}\right)}{\left(x^{2}-1\right)^{2}}$$

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

$$\frac{11x^{2}-11x^{0}-\left(11\times 2x^{1+1}+2x^{1}\right)}{\left(x^{2}-1\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{11x^{2}-11x^{0}-\left(22x^{2}+2x^{1}\right)}{\left(x^{2}-1\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{11x^{2}-11x^{0}-22x^{2}-2x^{1}}{\left(x^{2}-1\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(11-22\right)x^{2}-11x^{0}-2x^{1}}{\left(x^{2}-1\right)^{2}}$$

Tango 22 mai i 11.

$$\frac{-11x^{2}-11x^{0}-2x^{1}}{\left(x^{2}-1\right)^{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{-11x^{2}-11x^{0}-2x}{\left(x^{2}-1\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{-11x^{2}-11-2x}{\left(x^{2}-1\right)^{2}}$$