Question

Решите по алгебре пж

Answer 1

$456^{0}-(frac{1}{125})^{-frac{1}{3}} +6^{-2}=1-sqrt[3]{125}+frac{1}{6^{2}}=1-5+frac{1}{36}=-4+frac{1}{36}=-3frac{35}{36}\frac{12^{frac{2}{5}}}{4^{frac{2}{5}}*3^{frac{3}{5}}}=frac{4^{frac{2}{5}}*3^{frac{2}{5}}  }{4^{frac{2}{5}}*3^{frac{3}{5}}} =frac{3^{frac{2}{5}}}{3^{frac{3}{5}}}=3^{frac{2}{5}-frac{3}{5}}=3^{-frac{1}{5}}=frac{1}{sqrt[5]{3} }$

$2^{sqrt{3}-1}*2^{5-sqrt{3} }=2^{sqrt{3}-1+5-sqrt{3}}=2^{4}=16\sqrt[6]{frac{c^{5}b^{3}}{a} }*sqrt[6]{frac{cb^{3} }{a^{11}}} =sqrt[6]{frac{c^{5}b^{3}*cb^{3}}{a*a^{11}}}=sqrt[6]{frac{c^{6}b^{6}}{a^{12}}}=frac{bc}{a^{2}}\frac{(x^{2})^{5}}{x^{4} *x^{9}}=frac{x^{10} }{x^{13}}=x^{10-13}=x^{-3}=frac{1}{x^{3}}$

$sqrt[7]{(frac{3}{8})^{6}}<sqrt[7]{(frac{4}{9})^{6}}\sqrt[7]{(frac{27}{72})^{6}}<sqrt[7]{(frac{32}{72})^{6}}$