2/v5-2 Usuń niewymiernosc z mianownika reszta w załączniku

Rozwiązanie znajduje się w załączniku :)

[latex]a) frac{2}{sqrt5-2}=frac{2}{sqrt5-2}cdotfrac{sqrt5+2}{sqrt5+2}=frac{2(sqrt5+2)}{5-4}=2(sqrt5+2)=2sqrt5+4\ b) frac{12}{3-sqrt3}=frac{12}{3-sqrt3}cdotfrac{3+sqrt3}{3+sqrt3}=frac{12(3+sqrt3)}{9-3}=frac{12(3+sqrt3)}{6}=2(3+sqrt3)=\ =6+2sqrt6\ c) frac{2sqrt3}{2+sqrt6}=frac{2sqrt3}{2+sqrt6}cdotfrac{2-sqrt6}{2-sqrt6}=frac{2sqrt3(2-sqrt6)}{4-6}=frac{2sqrt3(2-sqrt6)}{-2}=\ =-sqrt3(2-sqrt6)=-2sqrt3+sqrt{18}=-2sqrt3+sqrt{9cdot2}=\ =-2sqrt3+3sqrt2[/latex] [latex]d) frac{2sqrt2}{2sqrt2+1}=frac{2sqrt2}{2sqrt2+1}cdotfrac{2sqrt2-1}{2sqrt2-1}=frac{2sqrt2(2sqrt2-1)}{8-1}=frac{8-2sqrt2}{7}\ e) frac{3sqrt2+3}{3sqrt2-3}=frac{3sqrt2+3}{3sqrt2-3}cdotfrac{3sqrt2+3}{3sqrt2+3}=frac{(3sqrt2+3)^2}{9cdot2-9}=frac{9cdot2+2cdot3cdot3sqrt2+9}{18-9}=\ =frac{18sqrt2+27}{9}=frac{9(2sqrt2+3)}{9}=2sqrt2+3\ [/latex] [latex]f) frac{4+2sqrt3}{4-2sqrt3}=frac{4+2sqrt3}{4-2sqrt3}cdotfrac{4+2sqrt3}{4+2sqrt3}=frac{(4+2sqrt3)^2}{16-4cdot3}=frac{16+2cdot4cdot2sqrt3+4cdot3}{16-12}=\ =frac{16+12+16sqrt3}{4}=frac{28+16sqrt3}{4}=frac{4(7+4sqrt3)}{4}=7+4sqrt3[/latex] [latex]g) frac{sqrt7-sqrt6}{sqrt7+sqrt6}=frac{sqrt7-sqrt6}{sqrt7+sqrt6}cdotfrac{sqrt7-sqrt6}{sqrt7-sqrt6}=frac{(sqrt7-sqrt6)^2}{7-6}=\ =(sqrt7-sqrt6)^2=(sqrt7)^2-2cdotsqrt7cdotsqrt6+(sqrt6)^2=\ =7-2sqrt{42}+6=13-2sqrt{42}[/latex] [latex]h) frac{sqrt3+sqrt2}{3sqrt2-2sqrt3}= frac{sqrt3+sqrt2}{3sqrt2-2sqrt3}cdot frac{3sqrt2+2sqrt3}{3sqrt2+2sqrt3}=frac{(sqrt3+sqrt2)(3sqrt2+2sqrt3)}{9cdot2-4cdot3}=\ =frac{3sqrt6+2cdot3+3cdot2+2sqrt6}{18-12}=frac{5sqrt6+12}{6}=frac{5sqrt6}{6}+2[/latex]