20. x = -√3 ... = (√(1 - x))² - x² = (1 - x) - x² = 1 - x - x² 1 - x - x²= 1 - (-√3) - (-√3)² = 1 + √3 - 3 = √3 - 2 ----- odpowiedź 4. ...= √(4^(1/3))* ∛(-2)^(1/2) * 8^(1/6) = 4^(1/3)^(1/2) * (-2)^(1/2)^(1/3) * 8^(1/6) = 4^(1/6) * (-2)^(1/6) * 8^(1/6) = [ 4 * (-2) * 8] ^(1/6) = (-64)^(1/6) =(-2^6)^(1/6) = (-2)^1 = -2 8. (a + b)² = a² + 2ab + b² ... =(3 - √2)+2 * √[ (3 - √2) * (3 + √2) ] + (3 + √2) = 3 - √2 + 2 * √[ 3² -(√2)²] + 3 + √2 = 6 + 2 * √(9 - 2) = 6 + 2 * √7 = 6 + 2√7
20.Korzystam ze wzoru skróconego mnozenia i własnosci: [latex](sqrt{a})^2=|a|[/latex] [latex](sqrt{1-x}-x)(sqrt{1-x}+x)=(sqrt{1-x})^2-x^2 = |1-x|-x^2[/latex] dla x = [latex]-sqrt{3}[/latex] : [latex]|1+sqrt{3}|-(-sqrt{3})^2=sqrt{3}+1-3 = sqrt{3}-2[/latex] 8.To zadanie opiera się na działaniu na potęgach :) [latex]sqrt{sqrt[3]{4}}cdotsqrt[3]{-sqrt{2}}cdotsqrt[6]{8}=(4^{frac{1}{3}})^{frac{1}{2}}cdot(-2^{frac{1}{2}})^{frac{1}{3}}cdot8^{frac{1}{6}}=4^{frac{1}{6}}cdot(-2)^{frac{1}{6}}cdot8^{frac{1}{6}}= \ =(4cdot(-2)cdot8)^{frac{1}{6}}=(-64)^{frac{1}{6}}=[(-2^6)]^{frac{1}{6}}=-2[/latex] 9. [latex](sqrt{3-sqrt{2}}+sqrt{3+sqrt{2}})^2 = \ =[(3-2^{frac{1}{2}})^{frac{1}{2}}]^2+2cdot(3-2^{frac{1}{2}})^{frac{1}{2}}cdot(3+2^{frac{1}{2}})^{frac{1}{2}}+[(3+2^{frac{1}{2}})^{frac{1}{2}}]^2= \ =3-2^{frac{1}{2}}+2cdotsqrt{(3-2^{frac{1}{2}})(3+2^{frac{1}{2}})}+3+2^{frac{1}{2}}= \ =6+2cdotsqrt{3^2-(2^{frac{1}{2}})^2}=6+2cdotsqrt{9-2}=6+2sqrt{7}=2(3+sqrt{7})[/latex] W tym zadaniu trzeba zamiast pierwiastka pisać liczbę do potęgi (1/2) no i zastosować wzór skróconego mnożenia :)
