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

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

Ovrednoti

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

Kratka navodila za rešitev

Če želite prišteti ali odšteti izraze, jih razširite na skupne imenovalce. Najmanjši skupni mnogokratnik 2\left(n+1\right) in 2n je 2n\left(n+1\right). Pomnožite \frac{2n^{2}-n-1}{2\left(n+1\right)} s/z \frac{n}{n}. Pomnožite \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} s/z \frac{n+1}{n+1}.

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

Ker \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} in \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} imata isti imenovalec, jih odštejte tako, da odštejete njihove števce.

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

Izvedi množenje v \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Združite podobne člene v 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Faktorizirajte izraze, ki še niso faktorizirani v \frac{2n^{2}+2n-2}{2n\left(n+1\right)}.

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

Okrajšaj 2 v števcu in imenovalcu.

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

Razčlenite n\left(n+1\right).

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

Če želite poiskati nasprotno vrednost za -\frac{1}{2}\sqrt{5}-\frac{1}{2}, poiščite nasprotno vrednost vsakega izraza.

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

Če želite poiskati nasprotno vrednost za \frac{1}{2}\sqrt{5}-\frac{1}{2}, poiščite nasprotno vrednost vsakega izraza.

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

Uporabite lastnost distributivnosti za množenje n+\frac{1}{2}\sqrt{5}+\frac{1}{2} krat n-\frac{1}{2}\sqrt{5}+\frac{1}{2} in kombiniranje pogojev podobnosti.

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

Kvadrat vrednosti \sqrt{5} je 5.

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

Pomnožite -\frac{1}{4} in 5, da dobite -\frac{5}{4}.

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

Seštejte -\frac{5}{4} in \frac{1}{4}, da dobite -1.

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

Razširi

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

Kratka navodila za rešitev

Če želite prišteti ali odšteti izraze, jih razširite na skupne imenovalce. Najmanjši skupni mnogokratnik 2\left(n+1\right) in 2n je 2n\left(n+1\right). Pomnožite \frac{2n^{2}-n-1}{2\left(n+1\right)} s/z \frac{n}{n}. Pomnožite \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} s/z \frac{n+1}{n+1}.

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

Ker \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} in \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} imata isti imenovalec, jih odštejte tako, da odštejete njihove števce.

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

Izvedi množenje v \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Združite podobne člene v 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Faktorizirajte izraze, ki še niso faktorizirani v \frac{2n^{2}+2n-2}{2n\left(n+1\right)}.

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

Okrajšaj 2 v števcu in imenovalcu.

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

Razčlenite n\left(n+1\right).

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

Če želite poiskati nasprotno vrednost za -\frac{1}{2}\sqrt{5}-\frac{1}{2}, poiščite nasprotno vrednost vsakega izraza.

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

Če želite poiskati nasprotno vrednost za \frac{1}{2}\sqrt{5}-\frac{1}{2}, poiščite nasprotno vrednost vsakega izraza.

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

Uporabite lastnost distributivnosti za množenje n+\frac{1}{2}\sqrt{5}+\frac{1}{2} krat n-\frac{1}{2}\sqrt{5}+\frac{1}{2} in kombiniranje pogojev podobnosti.

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

Kvadrat vrednosti \sqrt{5} je 5.

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

Pomnožite -\frac{1}{4} in 5, da dobite -\frac{5}{4}.

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

Seštejte -\frac{5}{4} in \frac{1}{4}, da dobite -1.

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