Question

Помогите плиз))запуталась

Answer 1

Объяснение:

$1)\ \frac{a-9a^{\frac{5}{6}} }{a^{\frac{1}{6}}-9 } .\\a^{\frac{1}{6}}=t\ \ \ \ \Rightarrow \\\frac{t^6-9t^5}{t-9} =\frac{t^5*(t-9)}{t-9}=t^5=(a^{\frac{1}{6}})^5 =a^{\frac{5}{6}} . \\2) \ \frac{4x^\frac{1}{4}-4x^\frac{1}{8}y^\frac{1}{6} +y^\frac{1}{3} }{2x^\frac{1}{4}-x^\frac{1}{8}y^\frac{1}{3} } .\\x^\frac{1}{8}=t\ \ \ \ \ \ y^\frac{1}{6}=v\ \ \ \ \Rightarrow$

$\frac{4t^2-4tv+v^2}{2t^2-tv^2}=\frac{(2t)^2-2*2t*v+v^2}{tv*(2t-v)} =\frac{(2t-v)^2}{tv*(2t-v)}=\frac{2t-v}{tv}=\frac{2x^\frac{1}{8}-y^\frac{1}{6} }{x^\frac{1}{8}y^\frac{1}{6} } .$

Answer 2

1 не решаю. Решено верно

2.

$\displaystyle \frac{4(x^{1/8})^2-4x^{1/8}*y^{1/6}+(y^{1/6})^2}{2(x^{1/8})^2*y^{1/6}-x^{1/8}*(y^{1/6})^2}}=\frac{(2x^{1/8}-y^{1/6})^2}{x^{1/8}*y^{1/6}*(2x^{1/8}-y^{1/6})}=\\\\=\frac{2x^{1/8}-y^{1/6}}{x^{1/8}*y^{1/6}}$

3.

$\displaystyle \frac{(a^{1/6})^2-(3b^{1/12})^2)}{a^{1/6}+3b^{1/12}}=a^{1/6}-3b^{1/12}}$