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

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

Procijeni

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

Kratki koraci rješenja

Da biste izvršili zbrajanje ili oduzimanje izraza, rastavite ih kako bi im nazivnici bili isti. Najmanji zajednički množilac brojeva 2\left(n+1\right) i 2n je 2n\left(n+1\right). Pomnožite \frac{2n^{2}-n-1}{2\left(n+1\right)} i \frac{n}{n}. Pomnožite \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} i \frac{n+1}{n+1}.

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

Pošto \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} i \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} imaju isti imenilac, oduzmite ih tako što ćete oduzeti njihove brojioce.

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

Izvršite množenja u \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Kombinirajte slične izraze u 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Faktorirajte izraze koji nisu već faktorirani u \frac{2n^{2}+2n-2}{2n\left(n+1\right)}.

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

Otkaži 2 u brojiocu i imeniocu.

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

Proširite n\left(n+1\right).

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

Da biste pronašli suprotnu vrijednost od -\frac{1}{2}\sqrt{5}-\frac{1}{2}, pronađite suprotnu vrijednost svakog izraza.

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

Da biste pronašli suprotnu vrijednost od \frac{1}{2}\sqrt{5}-\frac{1}{2}, pronađite suprotnu vrijednost svakog izraza.

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

Koristite svojstvo distributivnosti da biste pomnožili n+\frac{1}{2}\sqrt{5}+\frac{1}{2} s n-\frac{1}{2}\sqrt{5}+\frac{1}{2} i kombinirali slične pojmove.

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

Kvadrat broja \sqrt{5} je 5.

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

Pomnožite -\frac{1}{4} i 5 da biste dobili -\frac{5}{4}.

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

Saberite -\frac{5}{4} i \frac{1}{4} da biste dobili -1.

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

Proširi

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

Kratki koraci rješenja

Da biste izvršili zbrajanje ili oduzimanje izraza, rastavite ih kako bi im nazivnici bili isti. Najmanji zajednički množilac brojeva 2\left(n+1\right) i 2n je 2n\left(n+1\right). Pomnožite \frac{2n^{2}-n-1}{2\left(n+1\right)} i \frac{n}{n}. Pomnožite \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} i \frac{n+1}{n+1}.

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

Pošto \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} i \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} imaju isti imenilac, oduzmite ih tako što ćete oduzeti njihove brojioce.

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

Izvršite množenja u \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Kombinirajte slične izraze u 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Faktorirajte izraze koji nisu već faktorirani u \frac{2n^{2}+2n-2}{2n\left(n+1\right)}.

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

Otkaži 2 u brojiocu i imeniocu.

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

Proširite n\left(n+1\right).

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

Da biste pronašli suprotnu vrijednost od -\frac{1}{2}\sqrt{5}-\frac{1}{2}, pronađite suprotnu vrijednost svakog izraza.

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

Da biste pronašli suprotnu vrijednost od \frac{1}{2}\sqrt{5}-\frac{1}{2}, pronađite suprotnu vrijednost svakog izraza.

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

Koristite svojstvo distributivnosti da biste pomnožili n+\frac{1}{2}\sqrt{5}+\frac{1}{2} s n-\frac{1}{2}\sqrt{5}+\frac{1}{2} i kombinirali slične pojmove.

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

Kvadrat broja \sqrt{5} je 5.

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

Pomnožite -\frac{1}{4} i 5 da biste dobili -\frac{5}{4}.

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

Saberite -\frac{5}{4} i \frac{1}{4} da biste dobili -1.

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