Question

алгебра 1,2 помогите решить​

Answer 1

Ответ:

$1)\ \ \Big(\dfrac{125}{512}\Big)^{-\frac{1}{3} }=\Big(\dfrac{5^3}{8^3}\Big)^{-\frac{1}{3} }=\dfrac{5^{-1}}{8^{-1}}=\dfrac{8}{5}=1,6$

$2)\ \ x=0,008=\dfrac{8}{1000}=\dfrac{1}{125}=\dfrac{1}{5^3}\ \ \ ,\ \ \ \Big(\dfrac{1}{5^3}\Big)^{\frac{1}{3}}=\dfrac{1}{5}\ \ ,\\\\\dfrac{x^{\frac{1}{2}}-x^{\frac{5}{6}}}{x^{\frac{1}{2}}+x^{\frac{5}{6}}}=\dfrac{x^{\frac{1}{2}}\cdot (1-x^{\frac{1}{3}})}{x^{\frac{1}{2}}\cdot (1+x^{\frac{1}{3}})}=\dfrac{1-x^{\frac{1}{3}}}{1+x^{\frac{1}{3}}}=\dfrac{1-\dfrac{1}{5}}{1+\dfrac{1}{5}}=\dfrac{5-1}{5+1}=\dfrac{4}{6}=\dfrac{2}{3}$

Answer 2

$\displaystyle\bf\\1)\\\\\Big(\frac{125}{512} \Big)^{-\frac{1}{3} } =\Big(\frac{512}{125} \Big)^{\frac{1}{3} }=\Big[\Big(\frac{8}{5}\Big)^{3} \Big]^{\frac{1}{3} } =\frac{8}{5} =1,6 \\\\\\2)\\\\\frac{x^{\frac{1}{2} } -x^{\frac{5}{6} } }{x^{\frac{1}{2} }+x^{\frac{5}{6} }} =\frac{x^{\frac{1}{2} } (1-x^{\frac{1}{3} }) }{x^{\frac{1}{2} } (1+x^{\frac{1}{3} } )}=\frac{1-x^{\frac{1}{3} } }{1+x^{\frac{1}{3} } } \\\\x=0,008$

$\displaystyle\bf\\\frac{1-0,008^{\frac{1}{3} } }{1+0,008^{\frac{1}{3} } } =\frac{1-(0,2^{3} )^{\frac{1}{3} } }{1+(0,2^{3} )^{\frac{1}{3} } } =\frac{1-0,2}{1+0,2} =\frac{0,8}{1,2} =\frac{2}{3}$