Whakaoti i te 2n^2-n-1/2(n+1)-2(n-1)^2-(n-1)-1/2n

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

Aromātai

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

Ngā Hipanga Rongoā Poto

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

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

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

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

Mahia ngā whakarea i roto o \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Whakakotahitia ngā kupu rite i 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

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

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

Me whakakore tahi te 2 i te taurunga me te tauraro.

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

Whakarohaina te n\left(n+1\right).

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

Hei kimi i te tauaro o -\frac{1}{2}\sqrt{5}-\frac{1}{2}, kimihia te tauaro o ia taurangi.

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

Hei kimi i te tauaro o \frac{1}{2}\sqrt{5}-\frac{1}{2}, kimihia te tauaro o ia taurangi.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te n+\frac{1}{2}\sqrt{5}+\frac{1}{2} ki te n-\frac{1}{2}\sqrt{5}+\frac{1}{2} ka whakakotahi i ngā kupu rite.

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

Ko te pūrua o \sqrt{5} ko 5.

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

Whakareatia te -\frac{1}{4} ki te 5, ka -\frac{5}{4}.

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

Tāpirihia te -\frac{5}{4} ki te \frac{1}{4}, ka -1.

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

Whakaroha

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

Ngā Hipanga Rongoā Poto

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

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

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

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

Mahia ngā whakarea i roto o \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Whakakotahitia ngā kupu rite i 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

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

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

Me whakakore tahi te 2 i te taurunga me te tauraro.

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

Whakarohaina te n\left(n+1\right).

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

Hei kimi i te tauaro o -\frac{1}{2}\sqrt{5}-\frac{1}{2}, kimihia te tauaro o ia taurangi.

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

Hei kimi i te tauaro o \frac{1}{2}\sqrt{5}-\frac{1}{2}, kimihia te tauaro o ia taurangi.

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

Whakamahia te āhuatanga tuaritanga hei whakarea te n+\frac{1}{2}\sqrt{5}+\frac{1}{2} ki te n-\frac{1}{2}\sqrt{5}+\frac{1}{2} ka whakakotahi i ngā kupu rite.

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

Ko te pūrua o \sqrt{5} ko 5.

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

Whakareatia te -\frac{1}{4} ki te 5, ka -\frac{5}{4}.

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

Tāpirihia te -\frac{5}{4} ki te \frac{1}{4}, ka -1.

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