Réitigh 2n^2-n-1/2(n+1)-2(n-1)^2-(n-1)-1/2n

$$\frac { 2 n ^ { 2 } - n - 1 } { 2 ( n + 1 ) } - \frac { 2 ( n - 1 ) ^ { 2 } - ( n - 1 ) - 1 } { 2 n }$$

Luacháil

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

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

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

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

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

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

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

Cumaisc téarmaí comhchosúla in: 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

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

$$\frac{2\left(n-\left(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{2n\left(n+1\right)}$$

Cealaigh 2 mar uimhreoir agus ainmneoir.

$$\frac{\left(n-\left(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{n\left(n+1\right)}$$

Fairsingigh n\left(n+1\right)

$$\frac{\left(n-\left(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{n^{2}+n}$$

Chun an mhalairt ar -\frac{1}{2}\sqrt{5}-\frac{1}{2} a aimsiú, aimsigh an mhalairt ar gach téarma.

$$\frac{\left(n+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{n^{2}+n}$$

Chun an mhalairt ar \frac{1}{2}\sqrt{5}-\frac{1}{2} a aimsiú, aimsigh an mhalairt ar gach téarma.

$$\frac{\left(n+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(n-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)}{n^{2}+n}$$

Úsáid an t-airí dáileach chun n+\frac{1}{2}\sqrt{5}+\frac{1}{2} a mhéadú faoi n-\frac{1}{2}\sqrt{5}+\frac{1}{2} agus chun téarmaí comhchosúla a chumasc.

$$\frac{n^{2}+n-\frac{1}{4}\left(\sqrt{5}\right)^{2}+\frac{1}{4}}{n^{2}+n}$$

Is é 5 uimhir chearnach \sqrt{5}.

$$\frac{n^{2}+n-\frac{1}{4}\times 5+\frac{1}{4}}{n^{2}+n}$$

Méadaigh -\frac{1}{4} agus 5 chun -\frac{5}{4} a fháil.

$$\frac{n^{2}+n-\frac{5}{4}+\frac{1}{4}}{n^{2}+n}$$

Suimigh -\frac{5}{4} agus \frac{1}{4} chun -1 a fháil.

$$\frac{n^{2}+n-1}{n^{2}+n}$$

Fairsingigh

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

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

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

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

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

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

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

Cumaisc téarmaí comhchosúla in: 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

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

$$\frac{2\left(n-\left(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{2n\left(n+1\right)}$$

Cealaigh 2 mar uimhreoir agus ainmneoir.

$$\frac{\left(n-\left(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{n\left(n+1\right)}$$

Fairsingigh n\left(n+1\right)

$$\frac{\left(n-\left(-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{n^{2}+n}$$

Chun an mhalairt ar -\frac{1}{2}\sqrt{5}-\frac{1}{2} a aimsiú, aimsigh an mhalairt ar gach téarma.

$$\frac{\left(n+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(n-\left(\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\right)}{n^{2}+n}$$

Chun an mhalairt ar \frac{1}{2}\sqrt{5}-\frac{1}{2} a aimsiú, aimsigh an mhalairt ar gach téarma.

$$\frac{\left(n+\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\left(n-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)}{n^{2}+n}$$

Úsáid an t-airí dáileach chun n+\frac{1}{2}\sqrt{5}+\frac{1}{2} a mhéadú faoi n-\frac{1}{2}\sqrt{5}+\frac{1}{2} agus chun téarmaí comhchosúla a chumasc.

$$\frac{n^{2}+n-\frac{1}{4}\left(\sqrt{5}\right)^{2}+\frac{1}{4}}{n^{2}+n}$$

Is é 5 uimhir chearnach \sqrt{5}.

$$\frac{n^{2}+n-\frac{1}{4}\times 5+\frac{1}{4}}{n^{2}+n}$$

Méadaigh -\frac{1}{4} agus 5 chun -\frac{5}{4} a fháil.

$$\frac{n^{2}+n-\frac{5}{4}+\frac{1}{4}}{n^{2}+n}$$

Suimigh -\frac{5}{4} agus \frac{1}{4} chun -1 a fháil.

$$\frac{n^{2}+n-1}{n^{2}+n}$$