Question

Помогите с 2до 8номера дам 25б

Answer 1

$2); ; sqrt[3]{2sqrt3-4}cdot sqrt[3]{2sqrt3+4}=sqrt[3]{4cdot 3-16}=sqrt[3]{-2}=-sqrt[3]2approx -1,26\\-sqrt[3]{2}in [-2,-1]\\3); ; 5ab^2sqrt[3]{a^2b}=sqrt[3]{(5ab^2)^3cdot a^2b}=sqrt[3]{125a^3b^6cdot a^2b}=sqrt[3]{125a^5b^7}\\4); ; ysqrt[10]{ysqrt[3]{y^2}}:sqrt[6]{y^5}=ysqrt[10]{sqrt[3]{y^3cdot y^2}}:sqrt[6]{y^5}=\\=ysqrt[10]{sqrt[3]{y^5}}:sqrt[6]{y^5}=ysqrt[30]{y^5}:sqrt[6]{y^5}=ycdot sqrt[6]{y}:sqrt[6]{y^5}=$

$= frac{ysqrt[6]{y}}{sqrt[6]{y^5}} = sqrt[6]{frac{y^7}{y^5} }=sqrt[6]{y^2}=sqrt[3]{y}$

$5); ; a= frac{9}{16} ;\frac{a+sqrt{a}}{1+sqrt{a}} + frac{a-1}{1+sqrt{a} } = frac{2a+sqrt{a}-1}{1+sqrt{a}} = frac{frac{18}{16}+sqrt{frac{9}{16}}-1}{1+sqrt{frac{9}{16}}} = frac{frac{18}{16}+frac{3}{4}-1}{1+frac{3}{4}} = frac{frac{14}{16}}{frac{7}{4}} = \\=frac{14cdot 4}{16cdot 7} =frac{2}{4}=frac{1}{2}\\6); ; frac{sqrt{sqrt{c}+sqrt{d}}cdot sqrt{sqrt{c}-sqrt{d}}}{d-c} =frac{sqrt{c-d}}{-(c-d)} = -frac{1}{sqrt{c-d}}$

$7); ; sqrt{2msqrt[3]{frac{1}{4m^2}sqrt{frac{n}{m}}}}:sqrt[12]{nm}=sqrt{sqrt[3]{frac{8m^3}{4m^2}sqrt{frac{n}{m}}}}:sqrt[12]{nm}=\\=sqrt[6]{2msqrt{frac{n}{m}}}:sqrt[12]{nm}=sqrt[6]{sqrt{4m^2cdot frac{n}{m}}}:sqrt[12]{nm}=\\=sqrt[12]{4mn}cdot frac{1}{sqrt[12]{nm}}=sqrt[12]{frac{4mn}{nm}}=sqrt[12]{4}$

$8); ; t=3;\\frac{sqrt{t^3-8}}{sqrt{t-2}} = frac{sqrt{(t-2)(t^2+2t+4)}}{sqrt{t-2}} =sqrt{t^2+2t+4}=sqrt{9+6+4}=\\=sqrt{19}$