giúp tui câu e và d trong bài 1 và bài 2 a và b
Bài 1: $a)3\sqrt{18}-\sqrt{32}+4\sqrt2+\sqrt{162}$ $b)2\sqrt{48}-4\sqrt{27}+\sqrt{75}+\sqrt{12}c)\sqrt{(3-\sqrt7)^2}+\sqrt{(2-\sqrt7)^2}$,$d)\sqrt{21-8\sqrt5}-\sqrt{21+8\sqrt5}$ $e)3)\sqrt{4+\sqrt{5\sqrt3+5\sqrt{48-10\sqrt{7+4\sqrt3}}}}$,$ài2:a)\frac{\sqrt2}{\sqrt2-1}-\frac{\sqrt2}{\sqrt2+1}$ $b)\frac1{\sqrt{(\sqrt3+3)^2}}+\frac1{\sqrt{(\sqrt3-3)^2}}$

Question

giúp tui câu e và d trong bài 1 và bài 2 a và b
 Bài 1: $a)3\sqrt{18}-\sqrt{32}+4\sqrt2+\sqrt{162}$ $b)2\sqrt{48}-4\sqrt{27}+\sqrt{75}+\sqrt{12}c)\sqrt{(3-\sqrt7)^2}+\sqrt{(2-\sqrt7)^2}$,$d)\sqrt{21-8\sqrt5}-\sqrt{21+8\sqrt5}$ $e)3)\sqrt{4+\sqrt{5\sqrt3+5\sqrt{48-10\sqrt{7+4\sqrt3}}}}$,$ài2:a)\frac{\sqrt2}{\sqrt2-1}-\frac{\sqrt2}{\sqrt2+1}$ $b)\frac1{\sqrt{(\sqrt3+3)^2}}+\frac1{\sqrt{(\sqrt3-3)^2}}$

Accepted Answer

Bài 1:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
d,\sqrt{21-8\sqrt{5}} -\sqrt{21+8\sqrt{5}}\
=\sqrt{16-2.4.\sqrt{5} +5} -\sqrt{16+2.4.\sqrt{5} +5}\
=\sqrt{\left( 4-\sqrt{5} ight)^{2}} -\sqrt{\left( 4+\sqrt{5} ight)^{2}}\
=\mid 4-\sqrt{5} \mid -\mid 4+\sqrt{5} \mid \
=4-\sqrt{5} -4-\sqrt{5}\
=-2\sqrt{5}\
e,\sqrt{4+\sqrt{5\sqrt{3} +5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\
=\sqrt{4+\sqrt{5\sqrt{3} +5\sqrt{48-10\sqrt{\left( 2+\sqrt{3} ight)^{2}}}}}\
=\sqrt{4+\sqrt{5\sqrt{3} +5\sqrt{48-10\left( 2+\sqrt{3} ight)}}}\
=\sqrt{4+\sqrt{5\sqrt{3} +5\sqrt{28-10\sqrt{3}}}}\
=\sqrt{4+\sqrt{5\sqrt{3} +5\sqrt{\left( 5-\sqrt{3} ight)^{2}}}}\
=\sqrt{4+\sqrt{5\sqrt{3} +5\left( 5-\sqrt{3} ight)}}\
=\sqrt{4+\sqrt{25}}\
=\sqrt{4+5}\
=\sqrt{9} =3
\end{array}$
Bài 2:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
a,\frac{\sqrt{2}}{\sqrt{2} -1} -\frac{\sqrt{2}}{\sqrt{2} +1}\
=\frac{\sqrt{2}\left(\sqrt{2} +1 ight) -\sqrt{2}\left(\sqrt{2} -1 ight)}{\left(\sqrt{2} -1 ight)\left(\sqrt{2} +1 ight)}\
=2+\sqrt{2} -2+\sqrt{2}\
=2\sqrt{2}\
b,\frac{1}{\sqrt{\left(\sqrt{3} +3 ight)^{2}}} +\frac{1}{\sqrt{\left(\sqrt{3} -3 ight)^{2}}}\
=\frac{1}{\sqrt{3} +3} +\frac{1}{\mid \sqrt{3} -3\mid }\
=\frac{1}{\left(\sqrt{3} +3 ight)} +\frac{1}{3-\sqrt{3}}\
=\frac{3-\sqrt{3} +\sqrt{3} +3}{\left( 3+\sqrt{3} ight)\left( 3-\sqrt{3} ight)}\
=\frac{6}{9-3} =1
\end{array}$