solvefor x,f(x-2)=sqrt(x-3)

Lösung

löse nach

x, f (x - 2) = x - 3

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

Schritte zur Lösung

f (x - 2) = x - 3

Schreibe

f (x - 2)

um:

x f - 2 f

x f - 2 f = x - 3

Quadriere beide Seiten:

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3

x^{2} f^{2} - 4 x f^{2} + 4 f^{2} = x - 3

Löse

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

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

Überprüfe die Lösungen:

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

Die Lösungen sind

x = \frac{4 f ^{2} + 1 + - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}, x = \frac{4 f ^{2} + 1 - - 4 f ^{2} + 1}{2 f ^{2}} {0 < f \leq \frac{1}{2}}

x^{2} + 2 x = 27 - 4 x

5 x + 4 = (x + 4)^{\frac{1}{2}} + 2

2 x - 5 - 3 x + 4 = 1

x + 5 x = 12

188 - 818 = n