Đáp án:

`a)sqrt{(sqrt3-sqrt2)^2}+sqrt{(1-sqrt2)^2}=|sqrt3-sqrt2|+|1-sqrt2|=sqrt3-sqrt2+sqrt2-1=sqrt3-1` $\\$ `b)sqrt{(3-sqrt7)^2}+sqrt{(5-2sqrt7)^2}=|3-sqrt7|+|2sqrt7-5|=3-sqrt7+2sqrt7-5=sqrt7-2` $\\$ ` c)sqrt{3+2sqrt2}-sqrt{4-2sqrt3}=sqrt{2+2sqrt2+1}-sqrt{3-2sqrt3+1}=sqrt{(sqrt2+1)^2}-sqrt{(sqrt3-1)^2}=sqrt2+1-sqrt3+1=2+sqrt2-sqrt3` $\\$ `d)sqrt{12+6sqrt3}+sqrt{12-6sqrt3}=sqrt{9+2.3.sqrt3+3}+sqrt{9-2.3.sqrt3+3}=sqrt{(3+sqrt3)^2}+sqrt{(3-sqrt3)^2}=3+sqrt3+3-sqrt3=6` $\\$ `e)sqrt{9-4sqrt2}+sqrt{9+4sqrt2}=sqrt{8-2.2sqrt2+1}+sqrt{8+2.2sqrt2+1}=sqrt{(2sqrt2-1)^2}+sqrt{(2sqrt2+1)^2}=2sqrt2-1+2sqrt2+1=4sqrt2`

$a)\sqrt[]{(\sqrt[]{3}-\sqrt[]{2})^2}+\sqrt[]{(1-\sqrt[]{2})^2}$

$=|\sqrt[]{3}-\sqrt[]{2}|+|\sqrt[]{2}-1|$

$=\sqrt[]{3}-\sqrt[]{2}-1+\sqrt[]{2}$

$=\sqrt[]{3}-1$

$b)\sqrt[]{(3-\sqrt[]{7})^2}+\sqrt[]{(5-2\sqrt[]{7})^2}$

$=|3-\sqrt[]{7}|+|-5+2\sqrt[]{7}|$

$=3-5-\sqrt[]{7}+2\sqrt[]{7}$

$=-2+\sqrt[]{7}$

$c)\sqrt[]{3+2\sqrt[]{2}}-\sqrt[]{4-2\sqrt[]{3}}$

$=\sqrt[]{(\sqrt[]{2}+1)^2}-\sqrt[]{(\sqrt[]{3}-1)^2}$

$=|\sqrt[]{2}+1|-|\sqrt[]{3}-1|$

$=\sqrt[]{2}-\sqrt[]{3}+1+1$

$=\sqrt[]{2}-\sqrt[]{3}+2$

$d)\sqrt[]{12+6\sqrt[]{3}}+\sqrt[]{12-6\sqrt[]{3}}$

$=\sqrt[]{(3+\sqrt[]{3})^2}+\sqrt[]{(3-\sqrt[]{3})^2}$

$=|3+\sqrt[]{3}|+|3-\sqrt[]{3}|$

$=3+3+\sqrt[]{3}-\sqrt[]{3}$

$=6$

$e)\sqrt[]{9-4\sqrt[]{2}}+\sqrt[]{9+4\sqrt[]{2}}$

$=\sqrt[]{(2\sqrt[]{2}-1)^2}+\sqrt[]{(2\sqrt[]{2}+1)^2}$

$=|2\sqrt[]{2}-1|+|2\sqrt[]{2}+1|$

$=2\sqrt[]{2}+2\sqrt[]{2}-1+1$

$=4\sqrt[]{2}$