Question

Решите по алгебре пж

Answer 1

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

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

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