[tex]\it 1)\ \ x^2+15=51|_{-15}\Rightarrow x^2=36 \Rightarrow \sqrt{x^2}=\sqrt{36} \Rightarrow |x|=6 \Rightarrow x=\pm6\\ \\ 2)\ \ \dfrac{2}{3}x^2 +0,(6)=\sqrt{1\dfrac{7}{9}} \\ \\ \\ 0,(6)=\dfrac{\ 6^{(2}}{9} =\dfrac{2}{3}\\ \\ \\ \sqrt{1\dfrac{7}{9}}=\sqrt{\dfrac{16}{9}}=\dfrac{4}{3}[/tex]
Ecuația devine:
[tex]\it \dfrac{2}{3}x^2+\dfrac{2}{3}=\dfrac{4}{3}|_{\cdsot3} \Rightarrow 2x^2+2=4|_{-2} \Rightarrow 2x^2=2|_{:2} \Rightarrow x^2=1 \Rightarrow x=\pm1[/tex]
