[tex]\displaystyle a=\sqrt[3]{250}-\sqrt[3]{686}+\sqrt[3]{128}=5\sqrt[3]{2}-7\sqrt[3]{2}+4\sqrt[3]{2}=2\sqrt[3]{3}\\ \\ \\
b=(\frac{5}{\sqrt8+\sqrt3}+\frac 2{\sqrt3+1})\div(\frac {\sqrt6}{\sqrt3}+\sqrt2-\frac{\sqrt8}{2\sqrt2})=\\ \\
(\frac{5}{2\sqrt2+\sqrt3}+\frac 2{\sqrt3+1})\div(\sqrt2+\sqrt2-\frac{2\sqrt2}{2\sqrt2})=\\ \\
=\frac{5(\sqrt3+1)+2(2\sqrt2+\sqrt3)}{(2\sqrt2+\sqrt3)(\sqrt3+1)}\div(2\sqrt2-1)=\\ \\
=\frac{5\sqrt3+5+4\sqrt2+2\sqrt3}{(2\sqrt2+\sqrt3)(\sqrt3+1)(2\sqrt2-1)}=\\ \\[/tex]
[tex]\displaystyle =\frac{7\sqrt3+4\sqrt2+5}{(2\sqrt6+2\sqrt2+3+\sqrt3)(2\sqrt2-1)}=\\ \\
=\frac{7\sqrt3+4\sqrt2+5}{4\sqrt{12}-2\sqrt6+8-2\sqrt2+6\sqrt2-3+2\sqrt6-\sqrt3}=\\ \\
=\frac{7\sqrt3+4\sqrt2+5}{4\sqrt{12}+4\sqrt2-\sqrt3+5}=\\ \\
=\frac{7\sqrt3+4\sqrt2+5}{8\sqrt{3}+4\sqrt2-\sqrt3+5}=\\ \\
=\frac{7\sqrt3+4\sqrt2+5}{7\sqrt{3}+4\sqrt2+5}=1\\ \\ \\ \\
a)3*1-(2\sqrt[3]2)^3=3-16=-13,-13 =5(Fals)\\ \\
b)1=3\sqrt2 \sqrt[3]2(Fals)\\ \\
c)1\ \textless \ \sqrt[3]2(Adevarat)\\ \\
d)(\sqrt[3]2)^3=1^2(Fals)[/tex]
