Hejka prosze o pomoc w tych trzech zadankach ^^

1) [latex]=log_3162 - log_3( sqrt{2} )^2 = log_3( frac{162}{2})= log_381 = 4[/latex] 2) ⇔ 2x^2(2x -3) - 5(2x-3) = 0  ⇔  (2x^2 -5)(2x-3) = 0 ⇔  x = √5/2  lub x = -√5/2  lub  x = 3/2 3) = (20 - 8√5 +4)/ (3 - √5) = (24  - 8√5)/(3 - √5) = 8(3-√5)/(3-√5) = 8

[latex]25.\log_{3}162 - 2log_{3}sqrt{2}=log_{3}162 - log_{3}sqrt{2}^{2} = log_{3}162-log_{3}2=\\=log_{3}frac{162}{2} = log_{3}81 = log_{3}3^{4} = 4[/latex] [latex]26.\4x^{3}-6x^{2}-10x+15 = 0\\2x^{2}(2x-3)-5(2x-3) = 0\\(2x^{2}-5)(2x-3) = 0\\2(x +sqrt{frac{5}{2}})(x-sqrt{frac{5}{2}})(2x-3)= 0\\x=-frac{sqrt{5}}{sqrt{2}}cdotfrac{sqrt{2}}{sqrt{2}} = -frac{sqrt{10}}{2}\\lub\x = frac{sqrt{10}}{2}\\lub\x = frac{3}{2}\\x inlbrace-frac{sqrt{10}}{2};frac{3}{2};frac{sqrt{10}}{2} brace[/latex] [latex]27.\\m = frac{(2sqrt{5}-2)^{2}}{3-sqrt{5}}=frac{4cdot5-8sqrt{5}+4}{3-sqrt{5}}=frac{24-8sqrt{5}}{3-sqrt{5}}=frac{8(3-sqrt{5})}{3-sqrt{5}} = 8 in N[/latex]