Réitigh b^2+2/b^4-1+3/1-b^4

$$\frac { b ^ { 2 } + 2 } { b ^ { 4 } - 1 } + \frac { 3 } { 1 - b ^ { 4 } }$$

Luacháil

$$\frac{1}{b^{2}+1}$$

Céimeanna gearra réitigh

Fachtóirigh b^{4}-1. Fachtóirigh 1-b^{4}.

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

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(b-1\right)\left(b+1\right)\left(b^{2}+1\right) agus \left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right) ná \left(b-1\right)\left(b+1\right)\left(b^{2}+1\right). Méadaigh \frac{3}{\left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right)} faoi \frac{-1}{-1}.

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

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

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

Déan iolrúcháin in b^{2}+2+3\left(-1\right).

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

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

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

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

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

Cealaigh \left(b-1\right)\left(b+1\right) mar uimhreoir agus ainmneoir.

$$\frac{1}{b^{2}+1}$$

Fairsingigh

$$\frac{1}{b^{2}+1}$$

Céimeanna gearra réitigh

Fachtóirigh b^{4}-1. Fachtóirigh 1-b^{4}.

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

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(b-1\right)\left(b+1\right)\left(b^{2}+1\right) agus \left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right) ná \left(b-1\right)\left(b+1\right)\left(b^{2}+1\right). Méadaigh \frac{3}{\left(b-1\right)\left(b+1\right)\left(-b^{2}-1\right)} faoi \frac{-1}{-1}.

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

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

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

Déan iolrúcháin in b^{2}+2+3\left(-1\right).

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

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

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

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

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

Cealaigh \left(b-1\right)\left(b+1\right) mar uimhreoir agus ainmneoir.

$$\frac{1}{b^{2}+1}$$