Question

9*(4^1/x)+5*(6^1/x)<4*(9^1/x)

Answer 1

$9\cdot 4^{1/x}+5\cdot 6^{1/x}\ \textless \ 4\cdot 9^{1/x}\\\\9\cdot (2\cdot 2)^{1/x}+5\cdot (2\cdot 3)^{1/x}-4\cdot (3\cdot 3)^{1/x}\ \textless \ 0\; |:(3\cdot 3)^{1/x}\ \textgreater \ 0\\\\9\cdot (\frac{2}{3})^{2/x}+5\cdot ( \frac{2}{3} )^{1/x}-4\ \textless \ 0\\\\t=( \frac{2}{3} )^{1/x}\ \textgreater \ 0\; ,\quad 9t^2+5t-4\ \textless \ 0\\\\D=25+144=169\; ,\; \; t_1= \frac{-5-13}{18}=-1\\\\t_2=\frac{-5+13}{18} =\frac{8}{18}= \frac{4}{9} =(\frac{2}{3})^2\; \; \; \Rightarrow$

$+++(-1)---((\frac{2}{3})^2)+++$

$-1<t<(\frac{2}{3})^2\\\\0<(\frac{2}{3})^{1/x}<(\frac{2}{3})^2$

$Tak\; kak\; \; \frac{2}{3}<1\; ,\; to$

$\frac{1}{x}>2\; \; \; \Rightarrow \; \; \; \frac{1}{x}-2>0\\\\\frac{1-2x}{x}>0\; ,\; \; \frac{2x-1}{x}<0\; ,\\\\+++(0)---(\frac{1}{2})+++\\\\x\in (0,\frac{1}{2})$

Answer 2

Решение смотри на фото