Réitigh v/v+1+3/v-1-6/v^2-1

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

Luacháil

$$\frac{v+3}{v+1}$$

Céimeanna gearra réitigh

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de v+1 agus v-1 ná \left(v-1\right)\left(v+1\right). Méadaigh \frac{v}{v+1} faoi \frac{v-1}{v-1}. Méadaigh \frac{3}{v-1} faoi \frac{v+1}{v+1}.

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

Tá an t-ainmneoir céanna ag \frac{v\left(v-1\right)}{\left(v-1\right)\left(v+1\right)} agus \frac{3\left(v+1\right)}{\left(v-1\right)\left(v+1\right)} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

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

Déan iolrúcháin in v\left(v-1\right)+3\left(v+1\right).

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

Cumaisc téarmaí comhchosúla in: v^{2}-v+3v+3.

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

Fachtóirigh v^{2}-1.

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

Tá an t-ainmneoir céanna ag \frac{v^{2}+2v+3}{\left(v-1\right)\left(v+1\right)} agus \frac{6}{\left(v-1\right)\left(v+1\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

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

Cumaisc téarmaí comhchosúla in: v^{2}+2v+3-6.

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

Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana in \frac{v^{2}+2v-3}{\left(v-1\right)\left(v+1\right)}.

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

Cealaigh v-1 mar uimhreoir agus ainmneoir.

$$\frac{v+3}{v+1}$$

Difreálaigh w.r.t. v

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

Céimeanna ag Baint Úsáid as Riail na nDíorthach don Líon

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de v+1 agus v-1 ná \left(v-1\right)\left(v+1\right). Méadaigh \frac{v}{v+1} faoi \frac{v-1}{v-1}. Méadaigh \frac{3}{v-1} faoi \frac{v+1}{v+1}.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v\left(v-1\right)}{\left(v-1\right)\left(v+1\right)}+\frac{3\left(v+1\right)}{\left(v-1\right)\left(v+1\right)}-\frac{6}{v^{2}-1})$$

Tá an t-ainmneoir céanna ag \frac{v\left(v-1\right)}{\left(v-1\right)\left(v+1\right)} agus \frac{3\left(v+1\right)}{\left(v-1\right)\left(v+1\right)} agus, mar sin, is féidir iad a shuimiú trína n-uimhreoirí a shuimiú.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v\left(v-1\right)+3\left(v+1\right)}{\left(v-1\right)\left(v+1\right)}-\frac{6}{v^{2}-1})$$

Déan iolrúcháin in v\left(v-1\right)+3\left(v+1\right).

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v^{2}-v+3v+3}{\left(v-1\right)\left(v+1\right)}-\frac{6}{v^{2}-1})$$

Cumaisc téarmaí comhchosúla in: v^{2}-v+3v+3.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v^{2}+2v+3}{\left(v-1\right)\left(v+1\right)}-\frac{6}{v^{2}-1})$$

Fachtóirigh v^{2}-1.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v^{2}+2v+3}{\left(v-1\right)\left(v+1\right)}-\frac{6}{\left(v-1\right)\left(v+1\right)})$$

Tá an t-ainmneoir céanna ag \frac{v^{2}+2v+3}{\left(v-1\right)\left(v+1\right)} agus \frac{6}{\left(v-1\right)\left(v+1\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v^{2}+2v+3-6}{\left(v-1\right)\left(v+1\right)})$$

Cumaisc téarmaí comhchosúla in: v^{2}+2v+3-6.

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

Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana in \frac{v^{2}+2v-3}{\left(v-1\right)\left(v+1\right)}.

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

Cealaigh v-1 mar uimhreoir agus ainmneoir.

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

Do dhá fheidhm indifreáilte ar bith, is ionann díorthach líon an dá fheidhme agus an t-ainmneoir méadaithe faoi dhíorthach an uimhreora lúide an t-uimhreoir méadaithe faoi dhíorthach an ainmneora, agus iad ar fad roinnte faoin ainmneoir cearnaithe.

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

Is ionann díorthach iltéarmaigh agus suim dhíorthaigh a théarmaí. Is ionann díorthach téarma thairisigh agus 0. Is ionann díorthach ax^{n} agus nax^{n-1}.

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

Déan an uimhríocht.

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

Fairsingigh ag baint úsáid as an airí dáileach.

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

Chun cumhachtaí an bhoinn chéanna a mhéadú, suimigh a n-easpónaint.

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

Bain lúibíní ar bith nach bhfuil gá leo.

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

Cuir téarmaí cosúla le chéile.

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

Dealaigh 1 ó 1 agus 3 ó 1.

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

Do théarma ar bith t, t^{1}=t.

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

Do théarma ar bith t ach amháin 0, t^{0}=1.

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