n(x-3)=sqrt((x-3)-3)

Lösung

n (x - 3) = (x - 3) - 3

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}

Schritte zur Lösung

n (x - 3) = (x - 3) - 3

Schreibe

n (x - 3)

um:

n x - 3 n

n x - 3 n = x - 3 - 3

Quadriere beide Seiten:

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 6

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 6

Löse

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 6 : x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}}

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}}

Überprüfe die Lösungen:

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}

Die Lösungen sind

x = \frac{6 n ^{2} + 1 + - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}, x = \frac{6 n ^{2} + 1 - - 12 n ^{2} + 1}{2 n ^{2}} {0 < n < \frac{1}{2 3}}

x - 2 x - 1 = 0

1 = \frac{3 x + 2}{4}

a - 5 = 10

3 x^{2} + 2 = 0

7 x + 5 = x - 3