Réitigh v/v^2+17v+72-8/v^2+15v+56

$$\frac { v } { v ^ { 2 } + 17 v + 72 } - \frac { 8 } { v ^ { 2 } + 15 v + 56 }$$

Luacháil

$$\frac{v-9}{\left(v+7\right)\left(v+9\right)}$$

Céimeanna gearra réitigh

Fachtóirigh v^{2}+17v+72. Fachtóirigh v^{2}+15v+56.

$$\frac{v}{\left(v+8\right)\left(v+9\right)}-\frac{8}{\left(v+7\right)\left(v+8\right)}$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de \left(v+8\right)\left(v+9\right) agus \left(v+7\right)\left(v+8\right) ná \left(v+7\right)\left(v+8\right)\left(v+9\right). Méadaigh \frac{v}{\left(v+8\right)\left(v+9\right)} faoi \frac{v+7}{v+7}. Méadaigh \frac{8}{\left(v+7\right)\left(v+8\right)} faoi \frac{v+9}{v+9}.

$$\frac{v\left(v+7\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}-\frac{8\left(v+9\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}$$

Tá an t-ainmneoir céanna ag \frac{v\left(v+7\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)} agus \frac{8\left(v+9\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\frac{v\left(v+7\right)-8\left(v+9\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}$$

Déan iolrúcháin in v\left(v+7\right)-8\left(v+9\right).

$$\frac{v^{2}+7v-8v-72}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}$$

Cumaisc téarmaí comhchosúla in: v^{2}+7v-8v-72.

$$\frac{v^{2}-v-72}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}$$

Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana in \frac{v^{2}-v-72}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}.

$$\frac{\left(v-9\right)\left(v+8\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}$$

Cealaigh v+8 mar uimhreoir agus ainmneoir.

$$\frac{v-9}{\left(v+7\right)\left(v+9\right)}$$

Fairsingigh \left(v+7\right)\left(v+9\right)

$$\frac{v-9}{v^{2}+16v+63}$$

Difreálaigh w.r.t. v

$$\frac{207+18v-v^{2}}{v^{4}+32v^{3}+382v^{2}+2016v+3969}$$

Céimeanna ag Baint Úsáid as Riail na nDíorthach don Líon

Fachtóirigh v^{2}+17v+72. Fachtóirigh v^{2}+15v+56.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v}{\left(v+8\right)\left(v+9\right)}-\frac{8}{\left(v+7\right)\left(v+8\right)})$$

Chun cothromóidí a shuimiú nó a dhealú, fairsingigh iad chun a n-ainmneoirí a mheaitseáil. Is é an t-iolrach is lú coitianta de \left(v+8\right)\left(v+9\right) agus \left(v+7\right)\left(v+8\right) ná \left(v+7\right)\left(v+8\right)\left(v+9\right). Méadaigh \frac{v}{\left(v+8\right)\left(v+9\right)} faoi \frac{v+7}{v+7}. Méadaigh \frac{8}{\left(v+7\right)\left(v+8\right)} faoi \frac{v+9}{v+9}.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v\left(v+7\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}-\frac{8\left(v+9\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)})$$

Tá an t-ainmneoir céanna ag \frac{v\left(v+7\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)} agus \frac{8\left(v+9\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)} agus, mar sin, is féidir iad a dhealú trína n-uimhreoirí a dhealú.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v\left(v+7\right)-8\left(v+9\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)})$$

Déan iolrúcháin in v\left(v+7\right)-8\left(v+9\right).

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v^{2}+7v-8v-72}{\left(v+7\right)\left(v+8\right)\left(v+9\right)})$$

Cumaisc téarmaí comhchosúla in: v^{2}+7v-8v-72.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v^{2}-v-72}{\left(v+7\right)\left(v+8\right)\left(v+9\right)})$$

Fachtóirigh na sloinn nach bhfuil fachtóirithe cheana in \frac{v^{2}-v-72}{\left(v+7\right)\left(v+8\right)\left(v+9\right)}.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{\left(v-9\right)\left(v+8\right)}{\left(v+7\right)\left(v+8\right)\left(v+9\right)})$$

Cealaigh v+8 mar uimhreoir agus ainmneoir.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v-9}{\left(v+7\right)\left(v+9\right)})$$

Úsáid an t-airí dáileach chun v+7 a mhéadú faoi v+9 agus chun téarmaí comhchosúla a chumasc.

$$\frac{\mathrm{d}}{\mathrm{d}v}(\frac{v-9}{v^{2}+16v+63})$$

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.

$$\frac{\left(v^{2}+16v^{1}+63\right)\frac{\mathrm{d}}{\mathrm{d}v}(v^{1}-9)-\left(v^{1}-9\right)\frac{\mathrm{d}}{\mathrm{d}v}(v^{2}+16v^{1}+63)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

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}.

$$\frac{\left(v^{2}+16v^{1}+63\right)v^{1-1}-\left(v^{1}-9\right)\left(2v^{2-1}+16v^{1-1}\right)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

Simpligh.

$$\frac{\left(v^{2}+16v^{1}+63\right)v^{0}-\left(v^{1}-9\right)\left(2v^{1}+16v^{0}\right)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

Méadaigh v^{2}+16v^{1}+63 faoi v^{0}.

$$\frac{v^{2}v^{0}+16v^{1}v^{0}+63v^{0}-\left(v^{1}-9\right)\left(2v^{1}+16v^{0}\right)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

Méadaigh v^{1}-9 faoi 2v^{1}+16v^{0}.

$$\frac{v^{2}v^{0}+16v^{1}v^{0}+63v^{0}-\left(v^{1}\times 2v^{1}+v^{1}\times 16v^{0}-9\times 2v^{1}-9\times 16v^{0}\right)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

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

$$\frac{v^{2}+16v^{1}+63v^{0}-\left(2v^{1+1}+16v^{1}-9\times 2v^{1}-9\times 16v^{0}\right)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

Simpligh.

$$\frac{v^{2}+16v^{1}+63v^{0}-\left(2v^{2}+16v^{1}-18v^{1}-144v^{0}\right)}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

Cuir téarmaí cosúla le chéile.

$$\frac{-v^{2}+18v^{1}+207v^{0}}{\left(v^{2}+16v^{1}+63\right)^{2}}$$

Do théarma ar bith t, t^{1}=t.

$$\frac{-v^{2}+18v+207v^{0}}{\left(v^{2}+16v+63\right)^{2}}$$

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

$$\frac{-v^{2}+18v+207\times 1}{\left(v^{2}+16v+63\right)^{2}}$$

Do théarma ar bith t, t\times 1=t agus 1t=t.

$$\frac{-v^{2}+18v+207}{\left(v^{2}+16v+63\right)^{2}}$$