solvefor x,f(x-8)=sqrt(x+8)+2

Lösung

löse nach

x, f (x - 8) = x + 8 + 2

x = \frac{16 f ^{2} + 4 f + 1 + 64 f ^{2} + 8 f + 1}{2 f ^{2}}, x = \frac{16 f ^{2} + 4 f + 1 - 64 f ^{2} + 8 f + 1}{2 f ^{2}} {f \leq - \frac{1}{8}}

Schritte zur Lösung

f (x - 8) = x + 8 + 2

Schreibe

f (x - 8)

um:

x f - 8 f

x f - 8 f = x + 8 + 2

Quadratwurzeln entfernen

x^{2} f^{2} - 16 x f^{2} - 4 x f + 64 f^{2} + 32 f + 4 = x + 8

Löse

x^{2} f^{2} - 16 x f^{2} - 4 x f + 64 f^{2} + 32 f + 4 = x + 8 : x = \frac{16 f ^{2} + 4 f + 1 + 64 f ^{2} + 8 f + 1}{2 f ^{2}}, x = \frac{16 f ^{2} + 4 f + 1 - 64 f ^{2} + 8 f + 1}{2 f ^{2}}

x = \frac{16 f ^{2} + 4 f + 1 + 64 f ^{2} + 8 f + 1}{2 f ^{2}}, x = \frac{16 f ^{2} + 4 f + 1 - 64 f ^{2} + 8 f + 1}{2 f ^{2}}

Überprüfe die Lösungen:

x = \frac{16 f ^{2} + 4 f + 1 + 64 f ^{2} + 8 f + 1}{2 f ^{2}}

Wahr

, x = \frac{16 f ^{2} + 4 f + 1 - 64 f ^{2} + 8 f + 1}{2 f ^{2}} {f \leq - \frac{1}{8}}

Die Lösungen sind

x = \frac{16 f ^{2} + 4 f + 1 + 64 f ^{2} + 8 f + 1}{2 f ^{2}}, x = \frac{16 f ^{2} + 4 f + 1 - 64 f ^{2} + 8 f + 1}{2 f ^{2}} {f \leq - \frac{1}{8}}

6 - x = x - 6

x^{5} = x

40 + 3 x = x

2 x - 15 = \frac{x}{4}

\frac{8}{8 - 4 x} = 0