1)
a) Ta có:
$\dfrac{1}{2}\sqrt[]{120}=\dfrac{1}{2}.\sqrt[]{4.30}=\dfrac{1}{2}.2\sqrt[]{30}=\sqrt[]{30}$
$\dfrac{1}{3}\sqrt[]{270}=\dfrac{1}{3}.\sqrt[]{9.30}=\dfrac{1}{3}.3\sqrt[]{30}=\sqrt[]{30}$
Vậy $\dfrac{1}{2}\sqrt[]{120}=\dfrac{1}{3}\sqrt[]{270}$
b) $2\sqrt[]{\dfrac{1}{3}}=\dfrac{2\sqrt[]{3}}{3}$
Ta có: $(\sqrt[]{7})^2=7$
$(2\sqrt[]{3})^2=12$
Vì $7<12$ nên $\dfrac{1}{3}\sqrt[]{7}<2\sqrt[]{\dfrac{1}{3}}$
2)
a) $3\sqrt[]{2x}-4\sqrt[]{2x}+8-2\sqrt[]{x}$
$=-\sqrt[]{2}.\sqrt[]{x}+8-2\sqrt[]{x}$
$=8-\sqrt[]{x}(2+\sqrt[]{2})$