Ajutati-ma si pe mine, dau coroana !

O rezolv, ca model, pe prima. Restul le faci singur, dupa model.
(4x-5)/(x+3)<=1
(4x-5)/(x+3)-1<=0
(4x-5-x-3)/(x+3)<=0
(3x-8)/(x+3)<=0
Calculam semnul fiecareia din cele doua expresii si al raportului lor:
x                           -3      8/3
3x-8       -------------------- 0 +++++++++++++++
x+3        -------------0 ++++++++++++++++++++
(3x-8)/(x+3)  +++++/-------0++++++++++++++++
Deci x∈(-3; 8/3]
 

[tex]\displaystyle \mathtt{a) \frac{4x-5}{x+3} \leq 1 }\\ \\ \mathtt{ \frac{4x-5}{x+3} -1 \leq 0}\\ \\ \mathtt{ \frac{4x-5-(x+3)}{x+3} \leq 0 }\\ \\ \mathtt{ \frac{4x-5-x-3}{x+3} \leq 0 }\\ \\ \mathtt{ \frac{3x-8}{x+3} \leq 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~3x-8=0\Rightarrow x= \frac{8}{3}} \\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x+3=0\Rightarrow x=-3 }[/tex]

[tex]\displaystyle \mathtt{x~~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~~~~~-3~~~~~~~~~~~~~~~ \frac{8}{3}~~~~~~~~~~~~~~~~~~+\infty }\\ \\ \mathtt{3x-8~~~~~~~~~~~~~--------------~0++++++++}\\ \\ \mathtt{x+3~~~~~~~~~~~~~~~--------~0++++++++++++++}\\ \\ \mathtt{ \frac{3x-8}{x+3} ~~~~~~~~~~~~++++++++~~|~-----0++++++++}\\ \\ \\ \mathtt{x \in \left(-3; \frac{8}{3}\right] }[/tex]

[tex]\displaystyle \mathtt{b) \frac{2-3x}{x-1} \geq -2 }\\ \\ \mathtt{ \frac{2-3x}{x-1}+2 \geq 0 }\\ \\ \mathtt{ \frac{2-3x+2(x-1)}{x-1} \geq 0 }\\ \\ \mathtt{ \frac{2-3x+2x-2}{x-1} \geq 0}\\ \\ \mathtt{ \frac{-x}{x-1} \geq 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x=0 \Rightarrow x=0}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x-1=0 \Rightarrow x=1}[/tex]

[tex]\displaystyle \mathtt{x~~~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~~~~~~0~~~~~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~+\infty}\\ \\ \mathtt{-x~~~~~~~~~~~~~~~~~++++++++0---------------}\\ \\ \mathtt{x-1~~~~~~~~~~~~~~--------------~0+++++++++}\\ \\ \mathtt{ \frac{-x}{x-1} ~~~~~~~~~~~~~~--------0+++++~|~---------}\\ \\ \\ \mathtt{x \in [0;1)}[/tex]

[tex]\displaystyle \mathtt{c) \frac{2}{x-1}+ \frac{5x-3}{x-1} \geq 1} \\ \\ \mathtt{ \frac{2}{x-1}+ \frac{5x-3}{x-1}-1 \geq 0}\\ \\ \mathtt{ \frac{2+5x-3-(x-1)}{x-1} \geq 0 }\\ \\ \mathtt{ \frac{2+5x-3-x+1}{x-1} \geq 0 }\\ \\ \mathtt{ \frac{4x}{x-1} \geq 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~4x=0\Rightarrow x=0}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ x-1=0\Rightarrow x=1}[/tex]

[tex]\displaystyle \mathtt{x~~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~~~~~~0~~~~~~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~+\infty}\\ \\ \mathtt{4x~~~~~~~~~~~~~~~~~~-------~0+++++++++++++++}\\ \\ \mathtt{x-1~~~~~~~~~~~~~~--------------~0++++++++}\\ \\ \mathtt{ \frac{4x}{x-1}~~~~~~~~~~~~~+++++++~0------~|~++++++++ }\\ \\ \\ \mathtt{x \in (-\infty;0] \cup (1;+\infty)}[/tex]

[tex]\displaystyle \mathtt{d) \frac{1}{2(x+2)} - \frac{2x+1}{x+2}\ \textless \ \frac{1}{2}}\\ \\ \mathtt{ \frac{1}{2(x+2)}- \frac{2x+1}{x+2}- \frac{1}{2}\ \textless \ 0}\\ \\ \mathtt{ \frac{1-2(2x+1)-(x+2)}{2x+4}\ \textless \ 0 }\\ \\ \mathtt{ \frac{1-4x-2-x-2}{2x+4} \ \textless \ 0}\\ \\ \mathtt{ \frac{-5x-3}{2x+4}\ \textless \ 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-5x-3=0 \Rightarrow x=- \frac{3}{5}}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2x+4=0 \Rightarrow x=-2}[/tex]

[tex]\displaystyle \mathtt{x~~~~~~~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~~~-2~~~~~~~~~~~~~- \frac{3}{5}~~~~~~~~~~~~~~ + \infty }\\ \\ \mathtt{-5x-3~~~~~~~~~~~~~~~++++++++++++++~0-------}\\ \\ \mathtt{2x+4~~~~~~~~~~~~~~~~~--------~0+++++++++++++}\\ \\ \mathtt{ \frac{-5x-3}{2x+4} ~~~~~~~~~~~~~~--------~|~+++++~0-------}\\ \\ \\ \mathtt{x \in (-\infty;-2)\cup\left(- \frac{3}{5};+\infty\right) }[/tex]

[tex]\displaystyle \mathtt{e) \frac{(x+4)(x+1)}{2x+3} \ \textgreater \ 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x+4=0 \Rightarrow x=-4}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x+1=0 \Rightarrow x=-1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2x+3=0 \Rightarrow x=- \frac{3}{2} }[/tex]

[tex]\displaystyle \mathtt{x~~~~~~~~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~~~~-4~~~~~~~- \frac{3}{2}~~~~~~~~~-1 ~~~~~~~~+\infty}\\ \\ \mathtt{x+4~~~~~~~~~~~~~~~~~~~~-------0++++++++++++++}\\ \\ \mathtt{x+1~~~~~~~~~~~~~~~~~~~~----------------~0+++++}\\ \\ \mathtt{2x+3~~~~~~~~~~~~~~~~~~-----------~0++++++++++}\\ \\ \mathtt{ \frac{(x+4)(x+1)}{2x+3} ~~~~~~~~-------0+++~|~----~0+++++}\\ \\ \\ \mathtt{x \in \left(-4;- \frac{3}{2}\right) \cup (-1,+ \infty) }[/tex]

[tex]\displaystyle \mathtt{f) \frac{(-x-3)(x+5)}{x-2}\ \textless \ 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x-3=0 \Rightarrow x=-3}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x+5=0\Rightarrow x=-5}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x-2=0 \Rightarrow x=2}[/tex]

[tex]\displaystyle \mathtt{x~~~~~~~~~~~~~~~~~~~~~~~~~~~-\infty~~~~~~~~~-5~~~~~~~~~-3~~~~~~~~~~~2~~~~~~~~~~+\infty}\\ \\ \mathtt{-x-3~~~~~~~~~~~~~~~~~~~+++++++++++0----------}\\ \\ \mathtt{x+5~~~~~~~~~~~~~~~~~~~~~------~0+++++++++++++++}\\ \\\mathtt{x-2~~~~~~~~~~~~~~~~~~~~~~---------------~0+++++}\\ \\ \mathtt{ \frac{(-x-3)(x+5)}{x-2} ~~~~~~~~++++++0----~0++++~|~-----}\\ \\ \\ \mathtt{x \in (-5;-3) \cup(2;+ \infty)}[/tex]