Решите пожалуйста 4 неравенства с помощью систем , пользуйтесь пожалуйста тремя...

$1)\sqrt{3x-2}\ \textless \ x$
Данное неравенство равносильно системе трех неравенств:
3х-2 < x²
3x-2 > 0
x > 0

или

х²-3х+2>0
3x>2
x>0
   ///////////////     //////////////
-------(2/3)--(1)------(2)-----------
   \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Ответ.(2/3; 1) U (2; +∞)

$2)\sqrt{3x+4}\ \textgreater \ 2x+3$
$\left \{ {{3x+4\ \textgreater \ (2x+3)^2} \atop {2x+3\ \textgreater \ 0}} \right. \\ \\ \left \{ {{4x^2+9x+5\ \textless \ 0} \atop {2x\ \textgreater \ -3}} \right. \\ \\ \left \{ {{(4x+5)(x+1)\ \textless \ 0} \atop {x\ \textgreater \ -1,5}} \right.$
   +    - +
--(-1,5)--(-5/4)----(-1)--------
      ///////////////////////////////////
Ответ. (-5/4;-1)
$3) \sqrt[6]{x^2-2}\ \textless \ \sqrt[6]{4-x} \\ \\ \left \{ {{x^2-2\ \textless \ 4-x} \atop {x^2-2\ \textgreater \ 0}} \right. \\ \\ \left \{ {{x^2+x-6\ \textless \ 0} \atop {x^2-2\ \textgreater \ 0}} \right. \\ \\ \left \{ {{(x+3)(x-2)\ \textless \ 0} \atop {(x- \sqrt{2})(x+ \sqrt{2})\ \textgreater \ 0}} \right.$
   /////////////////     /////////////
---------(-3)---------(-√2)---------(√2)------(2)-----
Ответ. ( -3;-√2)U(√2; 2)

$4) log_3(x^3-x+24)\ \textgreater \ log_3(x^3+4x^2-5x)$

Логарифмическая функция с основанием 3 > 1 возрастает.
Неравенство равносильно системе неравенств:
$\left \{ {{x^3-x+24\ \textgreater \ x^3+4x^2-5x} \atop {x^3+4x^2-5x\ \textgreater \ 0}} \right. \\ \\ \left \{ {{4(x^2-x-6)\ \textless \ 0} \atop {x(x^2+4x-5)\ \textgreater \ 0}} \right. \\ \\ \left \{ {{4(x-3)(x+2)\ \textless \ 0} \atop {x(x-1)(x+5)\ \textgreater \ 0}} \right.$
///////////////////        ////////
--------------(-5)-----(-2)-----(0)---(1)------(3)----
   \\\\\\\\\\\\\\\\\\\\\\\\\\\
Ответ. (0;1)