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

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

Evalwa

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

Passi Qosra tas-Soluzzjonijiet

Biex iżżid jew tnaqqas l-espressjonijiet, espandihom biex id-denominaturi tagħhom ikunu l-istess. L-inqas multiplu komuni ta' 2\left(n+1\right) u 2n huwa 2n\left(n+1\right). Immultiplika \frac{2n^{2}-n-1}{2\left(n+1\right)} b'\frac{n}{n}. Immultiplika \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} b'\frac{n+1}{n+1}.

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

Billi \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} u \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} għandhom l-istess denominatur, naqqashom billi tnaqqas in-numeraturi tagħhom.

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

Agħmel il-multiplikazzjonijiet fi \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Ikkombina termini simili f'2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Iffattura l-espressjonijiet li mhumiex diġà fatturati f'\frac{2n^{2}+2n-2}{2n\left(n+1\right)}.

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

Annulla 2 fin-numeratur u d-denominatur.

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

Espandi n\left(n+1\right).

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

Biex issib l-oppost ta' -\frac{1}{2}\sqrt{5}-\frac{1}{2}, sib l-oppost ta' kull terminu.

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

Biex issib l-oppost ta' \frac{1}{2}\sqrt{5}-\frac{1}{2}, sib l-oppost ta' kull terminu.

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

Uża l-propjetà distributtiva biex timmultiplika n+\frac{1}{2}\sqrt{5}+\frac{1}{2} b'n-\frac{1}{2}\sqrt{5}+\frac{1}{2} u kkombina termini simili.

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

Il-kwadrat ta' \sqrt{5} huwa 5.

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

Immultiplika -\frac{1}{4} u 5 biex tikseb -\frac{5}{4}.

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

Żid -\frac{5}{4} u \frac{1}{4} biex tikseb -1.

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

Espandi

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

Passi Qosra tas-Soluzzjonijiet

Biex iżżid jew tnaqqas l-espressjonijiet, espandihom biex id-denominaturi tagħhom ikunu l-istess. L-inqas multiplu komuni ta' 2\left(n+1\right) u 2n huwa 2n\left(n+1\right). Immultiplika \frac{2n^{2}-n-1}{2\left(n+1\right)} b'\frac{n}{n}. Immultiplika \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} b'\frac{n+1}{n+1}.

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

Billi \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} u \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} għandhom l-istess denominatur, naqqashom billi tnaqqas in-numeraturi tagħhom.

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

Agħmel il-multiplikazzjonijiet fi \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Ikkombina termini simili f'2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Iffattura l-espressjonijiet li mhumiex diġà fatturati f'\frac{2n^{2}+2n-2}{2n\left(n+1\right)}.

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

Annulla 2 fin-numeratur u d-denominatur.

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

Espandi n\left(n+1\right).

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

Biex issib l-oppost ta' -\frac{1}{2}\sqrt{5}-\frac{1}{2}, sib l-oppost ta' kull terminu.

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

Biex issib l-oppost ta' \frac{1}{2}\sqrt{5}-\frac{1}{2}, sib l-oppost ta' kull terminu.

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

Uża l-propjetà distributtiva biex timmultiplika n+\frac{1}{2}\sqrt{5}+\frac{1}{2} b'n-\frac{1}{2}\sqrt{5}+\frac{1}{2} u kkombina termini simili.

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

Il-kwadrat ta' \sqrt{5} huwa 5.

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

Immultiplika -\frac{1}{4} u 5 biex tikseb -\frac{5}{4}.

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

Żid -\frac{5}{4} u \frac{1}{4} biex tikseb -1.

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