3y+7=5,x^7x-3y=1

Lösung

3 y + 7 = 5, x^{7} x - 3 y = 1

x = \frac{1}{2 ^{\frac{3}{4}} 2 - 1} + \frac{2 - 1}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 2 - 1} - \frac{2 - 1}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 1 + 2} + \frac{1 + 2}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 1 + 2} - \frac{1 + 2}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 1 + 2} + \frac{1 + 2}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 1 + 2} - \frac{1 + 2}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 2 - 1} + \frac{2 - 1}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 2 - 1} - \frac{2 - 1}{4 2 ^{3}} i, y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3}

Schritte zur Lösung

[3 y + 7 = 5 x^{7} x - 3 y = 1]

Löse

3 y + 7 = 5 : y = - \frac{2}{3}

Setze die Lösungen

y = - \frac{2}{3}

x^{7} x - 3 y = 1

ein

Für

x^{7} x - 3 y = 1

, ersetze

y

mit

- \frac{2}{3} : x = \frac{1}{2 ^{\frac{3}{4}} 2 - 1} + \frac{2 - 1}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 2 - 1} - \frac{2 - 1}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 1 + 2} + \frac{1 + 2}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 1 + 2} - \frac{1 + 2}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 1 + 2} + \frac{1 + 2}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 1 + 2} - \frac{1 + 2}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 2 - 1} + \frac{2 - 1}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 2 - 1} - \frac{2 - 1}{4 2 ^{3}} i

Verifiziere die Lösungen, in dem du sie in die Original-Gleichungen einsetzt.

Deshalb sind die finalen Lösungen für

3 y + 7 = 5, x^{7} x - 3 y = 1

x = \frac{1}{2 ^{\frac{3}{4}} 2 - 1} + \frac{2 - 1}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 2 - 1} - \frac{2 - 1}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 1 + 2} + \frac{1 + 2}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 1 + 2} - \frac{1 + 2}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 1 + 2} + \frac{1 + 2}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 1 + 2} - \frac{1 + 2}{4 2 ^{3}} i, x = - \frac{1}{2 ^{\frac{3}{4}} 2 - 1} + \frac{2 - 1}{4 2 ^{3}} i, x = \frac{1}{2 ^{\frac{3}{4}} 2 - 1} - \frac{2 - 1}{4 2 ^{3}} i, y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3} y = - \frac{2}{3}

\frac{3 x}{4} = 15

24 + 4 n = - (n + 6)

so l v e f or h, a = 2 π r^{2} + 2 π r h

x (x - 4) - 2 = x^{2} - 4 (x + 3)

- \frac{1}{6} \cdot x - \frac{1}{2} + 21 \cdot (x + 2) = \frac{- 6 \cdot ( x - 3 )}{7}