a [latex]4xsqrt{3}(frac{1}{8}sqrt{27}-0,75sqrt{12})=\frac{1}{8}4xsqrt{3}sqrt{27}-4xsqrt{3}frac{3}{4}sqrt{12}=\xfrac{1}{2}sqrt{3*27}-3xsqrt{3*12}=\9xfrac{1}{2}-18x=-frac{27}{2}x[/latex] b [latex]sqrt{3frac{1}{8}}(-4asqrt{18}-2sqrt{8}+8asqrt{frac{1}{2}}-2sqrt{50})=\sqrt{frac{25}{8}}(-4a3sqrt{2}-2*2sqrt{2}+8afrac{1}{2}sqrt{2}-2*5sqrt{2})=\sqrt{frac{25}{8}}(-12asqrt{2}-4sqrt{2}+4asqrt{2}-10sqrt{2})=\frac{5}{2sqrt{2}}(-8asqrt{2}-14sqrt{2})=\frac{5}{2sqrt{2}}(-8asqrt{2})-frac{5}{2sqrt{2}}14sqrt{2})=\5(-4a)-5*7=\-20a-35[/latex] c [latex](-5sqrt{5}+3ysqrt{10}+2sqrt{125}-9ysqrt{1frac{1}{9}})sqrt{frac{1}{5}}=\(-5sqrt{5}+3ysqrt{10}+2*5sqrt{5}-9ysqrt{frac{10}{9}})frac{1}{sqrt{5}}=\(-5sqrt{5}+3ysqrt{10}+10sqrt{5}-3ysqrt{10})frac{1}{sqrt{5}}=\(-5sqrt{5}+10sqrt{5})frac{1}{sqrt{5}}=\(5sqrt{5})frac{1}{sqrt{5}}=5[/latex] d [latex]2asqrt{2}(4asqrt{3}-frac{1}{2}sqrt{27}+asqrt{12})(-sqrt{2}sqrt{3})=\2asqrt{2}(4asqrt{3}-frac{1}{2}3sqrt{3}+2asqrt{3})(-sqrt{2}sqrt{3})=\-2a2sqrt{3}(4asqrt{3}-frac{1}{2}3sqrt{3}+2asqrt{3})=\-4asqrt{3}*4asqrt{3}-frac{1}{2}3sqrt{3}(-4asqrt{3})+2asqrt{3}(-4asqrt{3})=\-48a^{2}+18a-24a^{2}=\-72a^{2} + 18a ==== b √3i1/8 (-4a√18 - 2√8 +8a√1/2 - 2√50) = √25/8 (-4a3√2 - 2*2√2 + 8a 1/2 √2 - 2*5√2) = √(25/8) (-12a√2 - 4√2 + 4a√2 - 10√2) = 5/(2√2)(-8a√2 - 14√2) = 5/(2√2)*(-8a√2) - 5/(2√2)(14√2) = 5(-4a) - 5*7 = -20a - 35 c (-5√5 +3y√10 + 2√125 - 9y√1i1/9 ) √1/5= (-5√5 +3y√10 + 2*5√5 - 9y√10/√9 ) √1/5= (-5√5 +3y√10 + 10√5 - 3y√10 ) √1/5= (-5√5 + 10√5 ) √1/5= (5√5) √1/5= 5 d 2a√2 (4a√3 - 1/2 √27 + a√12)(-√6) = 2a√2 (4a√3 - 1/2 3√3 + 2a√3)(-√2√3) = -2a*2*√3 (4a√3 - 1/2 3√3 + 2a√3) = -2a*2*√3*4a√3 - 1/2 3√3(-2a*2*√3) + 2a√3*(-2a*2*√3) = -48a^2 + 18a - 24a^2 = -72a^2 + 18a
