[tex]\displaystyle\bf\\5)\\\\\sqrt[3]{x^{3} -2x+3} =x\\\\\\\Big(\sqrt[3]{x^{3} -2x+3} \Big)^{3} =x^{3} \\\\\\x^{3} -2x+3=x^{3} \\\\-2x+3=0\\\\-2x=-3\\\\\boxed{x=1,5}\\\\\\6)\\\\\sqrt{2x-1} =\sqrt{x^{2} +4x-16} \\\\\Big(\sqrt{2x-1} \Big)^{2} =\Big(\sqrt{x^{2} +4x-16} \Big)^{2} \\\\2x-1=x^{2} +4x-16\\\\x^{2} +2x-15=0\\\\Teorema \ Vieta \ : \\\\x_{1} + x_{2} =-2\\\\x_{1} \cdot x_{2} =-15\\\\x_{1} =3 \ \ , \ \ x_{2} =-5\\\\Proverka \ :\\1) \ x_{1} =3[/tex]

[tex]\displaystyle\bf\\\sqrt{2\cdot 3-1} =\sqrt{3^{2} +4\cdot 3-16} \\\\\sqrt{5} =\sqrt{5} - \ verno\\\\2) \ x_{2} =-5\\\\\sqrt{2\cdot(-5)-1}[/tex]

Под корнем квадратным отрицательное число , значит второй корень лишний .

Ответ : 3