👤

x=6([tex] \sqrt{x-2} -1 )..[/tex]...ajutor cv?:(

Răspuns :

[tex]x-6=\sqrt{(3x)}\\ (x-6)^2=3x\\ x^2-12x+36=3x\\ x^2-15x+36=0\\ (x-12)(x-3)=0\\ x=12, x=3.[/tex]
    
[tex]x=6( \sqrt{x-2} -1) \\ \\ \frac{x}{6}= \sqrt{x-2} -1 \\ \\ \frac{x}{6}+1= \sqrt{x-2} \\ \\ \text{Ridicam la puterea a 2-a si in stanga si in dreapta ecuatiei. } \\ \text{Ridicarea la putere poate introduce radacini false.} \\ \\ (\frac{x}{6}+1)^{2}= (\sqrt{x-2})^{2} \\ \\ (\frac{x+6}{6})^{2}= (\sqrt{x-2})^{2} \\ \\ [/tex]

[tex]\frac{x^{2}+12x+36}{36}= x-2 \\ x^{2}+12x+36=36(x-2) \\ x^{2}+12x+36=36x-72 \\ x^{2}+12x-36x+36+72=0 \\ x^{2}-24x+108 =0 \\ \\ x_{12}= \frac{24 \pm \sqrt{24^{2}-4*1*108} }{2}=\frac{24 \pm \sqrt{576-432} }{2} =\frac{24 \pm \sqrt{144} }{2}=\frac{24 \pm 12 }{2} \\ \\ x_{1}=\frac{24 + 12 }{2}= \frac{36}{2}=18 \\ \\ x_{2}=\frac{24 - 12 }{2}= \frac{12}{2}=6 [/tex]


[tex]\text{Ambele solutii verifica ecuatia initiala}[/tex]