[latex]cfrac{x + y}{sqrt{x^4 + 2x^3y + x^2y^2}} = cfrac{x + y}{sqrt{x^2(x^2 + 2xy + y^2)}} =[/latex] [latex]=cfrac{x + y}{sqrt{x^2(x + y)^2}} = cfrac{x+y}{|x| cdot |x + y|} =[/latex] [latex]=cfrac{x + y}{x(x + y)} = cfrac{1}{x}[/latex] [latex]sqrt{2x + 2sqrt{2x - 1}} - sqrt{2x - 2sqrt{2x - 1}} = m[/latex] [latex](2x + 2sqrt{2x - 1}) - 2 cdot sqrt{2x + 2sqrt{2x - 1}} cdot sqrt{2x - 2sqrt{2x - 1}} + (2x - 2sqrt{2x - 1}) = m^2[/latex] [latex]4x - 2 cdot sqrt{(2x + 2sqrt{2x-1})(2x-2sqrt{2x-1})} = m^2[/latex] [latex]4x - 2sqrt{4x^2 - 4(2x-1)} = m^2[/latex] [latex]4x - 2sqrt{4(x^2 - 2x + 1)} = m^2[/latex] [latex]4x - 2sqrt{4(x - 1)^2} = m^2[/latex] [latex]4x - 2 cdot 2(x - 1) = m^2[/latex] [latex]4x - 4x + 4 = m^2[/latex] [latex]m^2 = 4[/latex] [latex]m = 2 lor m = -2[/latex] Z faktu, że x > 1 wynika, że m > 0. Zatem [latex]sqrt{2x + 2sqrt{2x-1}} - sqrt{2x-2sqrt{2x-1}} = 2[/latex]
a)[latex] frac{x+y}{ sqrt{x^4+2x^3y+x^2y^2} } = frac{x+y}{ sqrt{(x^2+xy)^2} }= frac{x+y}{Ix^2+xyI} = frac{x+y}{x(x+y)} = frac{1}{x} [/latex] d)[latex] sqrt{2x+2 sqrt{2x-1} }- sqrt{2x-2 sqrt{2x-1} } [/latex] podnosimy wszystko do ² [latex]( sqrt{2x+2 sqrt{2x-1} }- sqrt{2x-2 sqrt{2x-1}})^2=[/latex]2x+2[latex] sqrt{2x-1} [/latex]-2[4x²-(2x-1)]+2x-2[latex] sqrt{2x-1} [/latex]=4x-8x²+16x+8=-8x²+20x+8
