oblicz wartość wyrażenia wymiernego: a)   [latex] frac{5x+2y}{x} dla x= -3 y=2[/latex]   b) [latex]frac{a^{2} - b^{2}}{a^{2} +2ab} dla a= sqrt{2} b= sqrt{3} [/latex]   c) [latex]frac{3x}{2x^{2}-5x} dla x= sqrt{3}+1 [/latex]     

a) jesli dobrze przepisalas zadanie to masz: x=-3y=2, czyli o x=2 y=-2/3 zatem (5*2-4/3)/2 = 5-2/3 = [latex]4frac{1}{3}[/latex] b) podstawiam: (2-3)/(2+2*sqrt(2*3)= -1/(2+sqrt(6)) *(2-sqrt(6))/(2-sqrt(6)) = -(2+sqrt(6))/(2^2+(2*sqrt(6))^2) = [-1(2-2*sqrt(6))] / (4-24) = (1-sqrt(6))/(10)   c) skracam x i mam: 3/(2x-5) = 3/(2sqrt3-3) * ((2sqrt3+3)/(2sqrt3+3)) = 3((2sqrt3+3)/(12-9) = 2sqrt3+3

a) [latex]frac{5x+2y}{x} =_{dla x= -3; y=2} =frac{5 cdot (-3)+2 cdot 2}{-3} = frac{-15+4}{-3}= frac{-11}{-3} = frac{11}{3} = 3frac{2}{3}[/latex]   b) [latex]frac{a^2- b^2}{a^2 +2ab} =_{dla a= sqrt{2}; b= sqrt{3}} =frac{(sqrt{2})^2- (sqrt{3})^2}{(sqrt{2})^2 +2 cdot sqrt{2} cdot sqrt{3}} = frac{2- 3}{2 +2 cdot sqrt{6}} = frac{-1}{2 +2 cdot sqrt{6}} = \\ = frac{-1 cdot (2 -2 cdot sqrt{6})}{(2 +2 cdot sqrt{6})(2 -2 cdot sqrt{6})} =frac{-1 cdot (2 -2 cdot sqrt{6})}{4-24} = =frac{-1 cdot (2 -2 cdot sqrt{6})}{-20} =frac{2 -2 cdot sqrt{6}}{20} = \\ =frac{2 cdot (1 -sqrt{6})}{20} =frac{1 -sqrt{6}}{10}[/latex]   c) [latex]frac{3x}{2x^2-5x}=_{dla x= sqrt{3}+1} =frac{3 cdot ( sqrt{3}+1)}{2 cdot ( sqrt{3}+1)^2-5 cdot ( sqrt{3}+1)}= frac{3sqrt{3}+3}{2 cdot (3+2 sqrt{3}+1)-5sqrt{3}-5}= \\ = frac{3sqrt{3}+3}{2 cdot (4+2 sqrt{3})-5sqrt{3}-5}= frac{3sqrt{3}+3}{8+4sqrt{3}-5sqrt{3}-5}= frac{3sqrt{3}+3}{3-sqrt{3}}=frac{(3sqrt{3}+3)(3+sqrt{3})}{(3-sqrt{3})(3+sqrt{3})}= \\ =frac{9sqrt{3}+9+9+3sqrt{3}}{9-3}=frac{12sqrt{3}+18}{6}=frac{6 cdot (2sqrt{3}+3)}{6}=2sqrt{3}+3[/latex]