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

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

Arvuta

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

Lühike lahenduskäik

Avaldiste liitmiseks või lahutamiseks laiendage need, et neil oleksid ühised nimetajad. 2\left(n+1\right) ja 2n vähim ühiskordne on 2n\left(n+1\right). Korrutage omavahel \frac{2n^{2}-n-1}{2\left(n+1\right)} ja \frac{n}{n}. Korrutage omavahel \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} ja \frac{n+1}{n+1}.

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

Kuna murdudel \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} ja \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} on sama nimetaja, lahutage nende lugejad.

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

Tehke korrutustehted võrrandis \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Kombineerige sarnased liikmed avaldises 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Kui avaldised pole tehtes \frac{2n^{2}+2n-2}{2n\left(n+1\right)} veel teguriteks lahutatud, tehke seda.

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

Taandage 2 nii lugejas kui ka nimetajas.

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

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

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

Avaldise "-\frac{1}{2}\sqrt{5}-\frac{1}{2}" vastandi leidmiseks tuleb leida iga liikme vastand.

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

Avaldise "\frac{1}{2}\sqrt{5}-\frac{1}{2}" vastandi leidmiseks tuleb leida iga liikme vastand.

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

Kasutage distributiivsusomadust, et korrutada n+\frac{1}{2}\sqrt{5}+\frac{1}{2} ja n-\frac{1}{2}\sqrt{5}+\frac{1}{2}, ning koondage sarnased liikmed.

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

\sqrt{5} ruut on 5.

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

Korrutage -\frac{1}{4} ja 5, et leida -\frac{5}{4}.

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

Liitke -\frac{5}{4} ja \frac{1}{4}, et leida -1.

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

Laienda

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

Lühike lahenduskäik

Avaldiste liitmiseks või lahutamiseks laiendage need, et neil oleksid ühised nimetajad. 2\left(n+1\right) ja 2n vähim ühiskordne on 2n\left(n+1\right). Korrutage omavahel \frac{2n^{2}-n-1}{2\left(n+1\right)} ja \frac{n}{n}. Korrutage omavahel \frac{2\left(n-1\right)^{2}-\left(n-1\right)-1}{2n} ja \frac{n+1}{n+1}.

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

Kuna murdudel \frac{\left(2n^{2}-n-1\right)n}{2n\left(n+1\right)} ja \frac{\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right)}{2n\left(n+1\right)} on sama nimetaja, lahutage nende lugejad.

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

Tehke korrutustehted võrrandis \left(2n^{2}-n-1\right)n-\left(2\left(n-1\right)^{2}-\left(n-1\right)-1\right)\left(n+1\right).

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

Kombineerige sarnased liikmed avaldises 2n^{3}-n^{2}-n-2n^{3}+2n^{2}+2n-2+n^{2}-1+n+1.

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

Kui avaldised pole tehtes \frac{2n^{2}+2n-2}{2n\left(n+1\right)} veel teguriteks lahutatud, tehke seda.

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

Taandage 2 nii lugejas kui ka nimetajas.

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

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

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

Avaldise "-\frac{1}{2}\sqrt{5}-\frac{1}{2}" vastandi leidmiseks tuleb leida iga liikme vastand.

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

Avaldise "\frac{1}{2}\sqrt{5}-\frac{1}{2}" vastandi leidmiseks tuleb leida iga liikme vastand.

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

Kasutage distributiivsusomadust, et korrutada n+\frac{1}{2}\sqrt{5}+\frac{1}{2} ja n-\frac{1}{2}\sqrt{5}+\frac{1}{2}, ning koondage sarnased liikmed.

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

\sqrt{5} ruut on 5.

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

Korrutage -\frac{1}{4} ja 5, et leida -\frac{5}{4}.

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

Liitke -\frac{5}{4} ja \frac{1}{4}, et leida -1.

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