Решите пожалуста !а),б),в),г),д),е)

Решите задачу:

$a)\;2\sqrt5-\sqrt{125}+3\sqrt{80}=2\sqrt5-\sqrt{25\cdot5}+3\sqrt{16\cdot5}=\\=2\sqrt5-5\sqrt5+12\sqrt5=9\sqrt5\\\\b)\;\sqrt6\cdot7\sqrt{54}=7\sqrt{6\cdot54}=7\sqrt{324}=7\cdot18=126\\\\c)\;(\sqrt{162}-10\sqrt2)\cdot\sqrt2=(\sqrt{81\cdot2}-10\sqrt2)\cdot\sqrt2=\\=(9\sqrt2-10\sqrt2)\cdot\sqrt2=(-\sqrt2)\cdot\sqrt2=-2\\\\d)\;(3\sqrt{98}-\sqrt{128}):\sqrt2=(3\sqrt{49\cdot2}-\sqrt{64\cdot2}):\sqrt2=\\=(21\sqrt2-8\sqrt2):\sqrt2=13\sqrt2:\sqrt2=13$

$e)\;4\sqrt{5t}-\frac34\sqrt{80t}-12\sqrt{\frac{125t}9}=4\sqrt{5}-\frac34\sqrt{16\cdot5t}-\frac{12}3\sqrt{25\cdot5t}=\\=4\sqrt{5t}-4\cdot\frac34\sqrt{5t}-4\cdot5\sqrt{5t}=4\sqrt{5t}-3\sqrt{5t}-20\sqrt{5t}=\\=-19\sqrt{5t}$