Whakaoti i te 3/1+x-2x^2+x/x-1

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

Aromātai

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

Ngā Hipanga Rongoā Poto

Tauwehea te 1+x-2x^{2}.

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

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

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

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

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

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

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

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{-3+2x^{2}+x}{\left(x-1\right)\left(2x+1\right)}.

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

Me whakakore tahi te x-1 i te taurunga me te tauraro.

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

Kimi Pārōnaki e ai ki x

$$-\frac{4}{\left(2x+1\right)^{2}}$$

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

Tauwehea te 1+x-2x^{2}.

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

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

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

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

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

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

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

Me whakatauwehe ngā kīanga kāore anō i whakatauwehea i roto o \frac{-3+2x^{2}+x}{\left(x-1\right)\left(2x+1\right)}.

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

Me whakakore tahi te x-1 i te taurunga me te tauraro.

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

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.

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

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}.

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

Mahia ngā tātaitanga.

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

Whakarohaina mā te āhuatanga tohatoha.

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

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

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

Mahia ngā tātaitanga.

$$\frac{4x^{1}+2x^{0}-\left(4x^{1}+6x^{0}\right)}{\left(2x^{1}+1\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{4x^{1}+2x^{0}-4x^{1}-6x^{0}}{\left(2x^{1}+1\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(4-4\right)x^{1}+\left(2-6\right)x^{0}}{\left(2x^{1}+1\right)^{2}}$$

Tangohia te 4 i 4 me te 6 i te 2.

$$\frac{-4x^{0}}{\left(2x^{1}+1\right)^{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{-4x^{0}}{\left(2x+1\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{-4}{\left(2x+1\right)^{2}}$$