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Calculați
[tex] \sqrt{9x^2-6x+1} + \sqrt{9x^2+6x+1} [/tex], pentru |x| [tex] \leq [/tex] 0,(3).


Răspuns :

Vezi abordarea in atasament. Distractie!
Vezi imaginea АНОНИМ
[tex] \sqrt{ 9x^{2} -6x+1} + \sqrt{ 9x^{2} +6x+1} = \\= \sqrt{ (3x-1)^{2}} + \sqrt{ (3x+1)^{2}} = \\= |3x-1|+|3x+1| = \\= |3x-1|+|3x+1| \\ |x|\ \textless \ = \frac{1}{3} \\ \frac{-1}{3} \ \textless \ =x\ \textless \ =\frac{1}{3} \\ (3x+1)\ \textgreater \ =0 ..oricare..ar..fi..x \\ (3x-1)\ \textless \ =0..oricare..ar..fi..x \\ |3x-1|+|3x+1|= \\ =1-3x+3x+1=\\=2[/tex]