Решите пжжжжж. СРОЧНО!!!

Решите задачу:

$\sqrt{2}( \sqrt{25*2}+ \sqrt{4*2})= \sqrt{2}(5 \sqrt{2}+2 \sqrt{2})= \sqrt{2}*7 \sqrt{2}=7*2=14$

$( \sqrt{3}- \sqrt{4*3})* \sqrt{3} =( \sqrt{3}-2 \sqrt{3}) \sqrt{3}=- \sqrt{3} \sqrt{3}=-3$

$\sqrt{7}( \sqrt{7}- \sqrt{4*7})= \sqrt{7}( \sqrt{7} -2 \sqrt{7} )= -\sqrt{7} \sqrt{7}=-7$

$( \sqrt{9*2}+ \sqrt{36*2}) \sqrt{2}=(3 \sqrt{2}+6 \sqrt{2})=9 \sqrt{2} \sqrt{2}=18$

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

$2*2+2 \sqrt{2} \sqrt{5} - \sqrt{2} \sqrt{5}-5= \sqrt{2} \sqrt{5}-1$

$a^2+a \sqrt{b}-a \sqrt{b}-b=a^2-b$

$b- \sqrt{c} \sqrt{b}+ \sqrt{b} \sqrt{c}-c=b-c$

$16+4 \sqrt{3}-4 \sqrt{3}-3=13$