Помогите решить по алгебре

$\sqrt{2} * \sqrt{32} = \sqrt{2*32} = \sqrt{64} = 8 \\ \sqrt{113^2-112^2} = \sqrt{ (113-112)(113+112) } = \sqrt{225} = 15 \\ \frac{ \sqrt{216} }{ \sqrt{6} } = \sqrt{216/6} = \sqrt{36} =6\\ \sqrt{5^4 * 3^2} = 5^2 * 3 = 25*3 = 75\\$

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$4 \sqrt{20} - \sqrt{125} = 4* \sqrt{4*5} - \sqrt{5*25} = 8 \sqrt{5} -5 \sqrt{5} =3 \sqrt{5}$

$(3 \sqrt{6} + \sqrt{12} ) \sqrt{3} =3 \sqrt{6*3} + \sqrt{12*3} =3 \sqrt{9*2} + \sqrt{36} = 9 \sqrt{2}+6$

$(5- \sqrt{2} )^2 = 25 - 10 \sqrt{2} + 2 = 23 - 10 \sqrt{2}$

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$12 \sqrt{3} = \sqrt{12^2 * 3} = \sqrt{144*3} = \sqrt{432} \\ 2 \sqrt{2} = \sqrt{2^2*2} = \sqrt{8} \\$