Найдите множество решений неравенства а) [latex] x^{ \frac{1}{2} } -7x^{ \frac{1}{4}} +6\ \textless \ 0[/latex] б) [latex] \frac{ x^{ \frac{4}{3} }-4 }{ x^{ \frac{2}{3} } +2 } - \frac{ x^{ \frac{2}{3} }-1 }{ x^{ \frac{1}{3} }-1...

[latex]a)\; x^{\frac{1}{2}}-7x^{\frac{1}{4}}+6<0\\\\t=x^{\frac{1}{4}},\; \; t^2-7t+6<0\\\\t_1=1,\; t_2=6\\\\+++(1)---(6)+++\\\\ \left \{ {{t>1} \atop {t<6}} \right. \; \left \{ {{x^{\frac{1}{4}}>1}} \atop {x^{\frac{1}{4}}<6}}} \right. \; \left \{ {{x>1} \atop {x<6^4}} \right. \; x\in (1,1296)\\\\b)\; \frac{x^{\frac{4}{3}}-4}{x^{\frac{2}{3}}+2}-\frac{x^{\frac{2}{3}}-1}{x^{\frac{1}{3}}-1}<3,\; x\ne 1\\\\\frac{(x^{\frac{2}{3}}-2)(x^{\frac{2}{3}}+2)}{x^{\frac{2}{3}}+2}-\frac{(x^{\frac{1}{3}}-1)(x^{\frac{1}{3}}+1)}{x^{\frac{1}{3}}-1}<3\\\\x^{\frac{2}{3}}-2-x^{\frac{1}{3}}-1<3\\\\t=x^{\frac{1}{3}},t^2-t-6<0\\\\-2-2,x>(-2)^3,x>-8\\\\x^{\frac{1}{3}}<3,x<3^3,x<27\\\\x\in (-8,1)U(1,27)[/latex]

а) √x -  7x^(1/4) +6 <0; делаем замену  : t = x^(1/4) , получим : (t -1)(t-6) < 0 ; 1