Riješi z+1/4=1/4n^2+2/3+3-[1/4(n-1)^2+frac{2}{3}(n-1)+3]

$$z + \frac { 1 } { 4 } = \frac { 1 } { 4 } n ^ { 2 } + \frac { 2 } { 3 } + 3 - [ \frac { 1 } { 4 } ( n - 1 ) ^ { 2 } + \frac { 2 } { 3 } ( n - 1 ) + 3 ]$$

Izračunaj n

$$n=5-6z$$

Koraci rješavanja linearne jednadžbe

Dodajte \frac{2}{3} broju 3 da biste dobili \frac{11}{3}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n-1\right)^{2}+\frac{2}{3}\left(n-1\right)+3\right)$$

Upotrijebite binomni teorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} da biste proširili \left(n-1\right)^{2}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n^{2}-2n+1\right)+\frac{2}{3}\left(n-1\right)+3\right)$$

Koristite svojstvo distributivnosti da biste pomnožili \frac{1}{4} s n^{2}-2n+1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}\left(n-1\right)+3\right)$$

Koristite svojstvo distributivnosti da biste pomnožili \frac{2}{3} s n-1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}n-\frac{2}{3}+3\right)$$

Kombinirajte -\frac{1}{2}n i \frac{2}{3}n da biste dobili \frac{1}{6}n.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{1}{4}-\frac{2}{3}+3\right)$$

Oduzmite \frac{2}{3} od \frac{1}{4} da biste dobili -\frac{5}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n-\frac{5}{12}+3\right)$$

Dodajte -\frac{5}{12} broju 3 da biste dobili \frac{31}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}\right)$$

Da biste pronašli suprotnu vrijednost izraza \frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}, pronađite suprotnu verziju svakog člana.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\frac{1}{4}n^{2}-\frac{1}{6}n-\frac{31}{12}$$

Kombinirajte \frac{1}{4}n^{2} i -\frac{1}{4}n^{2} da biste dobili 0.

$$z+\frac{1}{4}=\frac{11}{3}-\frac{1}{6}n-\frac{31}{12}$$

Oduzmite \frac{31}{12} od \frac{11}{3} da biste dobili \frac{13}{12}.

$$z+\frac{1}{4}=\frac{13}{12}-\frac{1}{6}n$$

Zamijenite strane tako da svi izrazi s nepoznanicama budu s lijeve strane.

$$\frac{13}{12}-\frac{1}{6}n=z+\frac{1}{4}$$

Oduzmite \frac{13}{12} od obiju strana.

$$-\frac{1}{6}n=z+\frac{1}{4}-\frac{13}{12}$$

Oduzmite \frac{13}{12} od \frac{1}{4} da biste dobili -\frac{5}{6}.

$$-\frac{1}{6}n=z-\frac{5}{6}$$

Pomnožite obje strane s -6.

$$\frac{-\frac{1}{6}n}{-\frac{1}{6}}=\frac{z-\frac{5}{6}}{-\frac{1}{6}}$$

Dijeljenjem s -\frac{1}{6} poništava se množenje s -\frac{1}{6}.

$$n=\frac{z-\frac{5}{6}}{-\frac{1}{6}}$$

Podijelite z-\frac{5}{6} s -\frac{1}{6} tako da pomnožite z-\frac{5}{6} s brojem recipročnim broju -\frac{1}{6}.

$$n=5-6z$$

Izračunaj z

$$z=\frac{5-n}{6}$$

Koraci rješenja

Dodajte \frac{2}{3} broju 3 da biste dobili \frac{11}{3}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n-1\right)^{2}+\frac{2}{3}\left(n-1\right)+3\right)$$

Upotrijebite binomni teorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} da biste proširili \left(n-1\right)^{2}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}\left(n^{2}-2n+1\right)+\frac{2}{3}\left(n-1\right)+3\right)$$

Koristite svojstvo distributivnosti da biste pomnožili \frac{1}{4} s n^{2}-2n+1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}\left(n-1\right)+3\right)$$

Koristite svojstvo distributivnosti da biste pomnožili \frac{2}{3} s n-1.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}-\frac{1}{2}n+\frac{1}{4}+\frac{2}{3}n-\frac{2}{3}+3\right)$$

Kombinirajte -\frac{1}{2}n i \frac{2}{3}n da biste dobili \frac{1}{6}n.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{1}{4}-\frac{2}{3}+3\right)$$

Oduzmite \frac{2}{3} od \frac{1}{4} da biste dobili -\frac{5}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n-\frac{5}{12}+3\right)$$

Dodajte -\frac{5}{12} broju 3 da biste dobili \frac{31}{12}.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\left(\frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}\right)$$

Da biste pronašli suprotnu vrijednost izraza \frac{1}{4}n^{2}+\frac{1}{6}n+\frac{31}{12}, pronađite suprotnu verziju svakog člana.

$$z+\frac{1}{4}=\frac{1}{4}n^{2}+\frac{11}{3}-\frac{1}{4}n^{2}-\frac{1}{6}n-\frac{31}{12}$$

Kombinirajte \frac{1}{4}n^{2} i -\frac{1}{4}n^{2} da biste dobili 0.

$$z+\frac{1}{4}=\frac{11}{3}-\frac{1}{6}n-\frac{31}{12}$$

Oduzmite \frac{31}{12} od \frac{11}{3} da biste dobili \frac{13}{12}.

$$z+\frac{1}{4}=\frac{13}{12}-\frac{1}{6}n$$

Oduzmite \frac{1}{4} od obiju strana.

$$z=\frac{13}{12}-\frac{1}{6}n-\frac{1}{4}$$

Oduzmite \frac{1}{4} od \frac{13}{12} da biste dobili \frac{5}{6}.

$$z=\frac{5}{6}-\frac{1}{6}n$$