[tex]5)\\\sqrt{x}+2x=1\\t=\sqrt{x}\Rightarrow t\geq0\\2t^2 +t=1\\t^2+\frac{1}{2}t=\frac{1}{2}\\(t+\frac{1}{4})^2=\frac{8}{16}+\frac{1}{16}\\t=\pm\frac{3}{4}-\frac{1}{4}\\t\geq0, \ wiec \ \ t=\frac{1}{2}\\\sqrt{x}=\frac{1}{2}\\x=\frac{1}{4}\\\\6)\\\sqrt[3]{x^2} -9\sqrt[3]{x}+8=0\\t=\sqrt[3]{x}\\t^2-9t=-8\\(t-\frac{9}{2})^2= \frac{-32}{4}+\frac{81}{4}=\frac{49}{4}\\t=\pm\frac{7}{2}+\frac{9}{2}\\t\in\{1,8\}\\\sqrt[3]{x}=1 \ \vee \ \sqrt[3]{x}=8\\x=1 \ \vee \ x=8^3\\x\in\{1,8^3\}\\\\[/tex]
[tex]\\\\7)\\x^6+x^3-2=0\\t=x^3\\t^2+t=2\\(t+\frac{1}{2})^2=\frac{8}{4}+\frac{1}{4}=\frac{9}{4}\\t=\pm\frac{3}{2}-\frac{1}{2}\\t\in\{-2,1\}\\x^3=-2 \ \vee \ x^3=1\\x=\sqrt[3]{-2} \ \vee \ x=1\\x\in\{\sqrt[3]{-2},1\}[/tex]
