Bonsoir,
1) [tex]\dfrac{x+3}{2-x}<1\\\\\dfrac{x+3}{2-x}-1<0\\\\\dfrac{(x+3)-(2-x)}{2-x}<0\\\\\dfrac{x+3-2+x}{2-x}<0\\\\\dfrac{2x+1}{2-x}<0 [/tex]
Tableau de signes.
Racines : 2x + 1 = 0 ==> 2x = -1
==> x = -1/2
2 - x =0 ==> -x = -2
==> x = 2
[tex]\begin{array}{|c|ccccccc|}x&-\infty&&-\frac{1}{2}&&2&&+\infty\\2x+1&&-&0&+&+&+& \\2-x&&+&+&+&0&-& \\ (2x+1)(2-x)&&-&0&+&|&-&\\ \end{array}\\\\\\S=]-\infty;-\dfrac{1}{2}[\ \cup\ ]2;+\infty[[/tex]
2) [tex]\dfrac{9x-4x^2}{2-3x}>0\\\\\dfrac{x(9-4x)}{2-3x}>0[/tex]
Tableau de signes.
Racines : x = 0
9 - 4x 0 ==> 4x = 9
==> x = 9/4
2 - 3x =0 ==> 3x = 2
==> x = 2/3
[tex]\begin{array}{|c|ccccccccc|}x&-\infty&&0&&\frac{2}{3}&&\frac{9}{4}&&+\infty\\x&&-&0&+&+&+&+&+& \\9-4x&&+&+&+&+&+&0&-& \\2-3x&&+&+&+&0&-&-&-& \\ \dfrac{x(9-4x)}{2-3x}&&-&0&+&|&-&0&+\\ \end{array}\\\\\\S=]0;\dfrac{2}{3}[\ \cup\ ]\dfrac{9}{4};+\infty[[/tex]
3) [tex](x+2)^2-(x+2)(3x+4)\ge0\\(x+2)(x+2)-(x+2)(3x+4)\ge0\\(x+2)[(x+2)-(3x+4)]\ge0\\(x+2)(x+2-3x-4)\ge0 \\(x+2)(-x-2)\ge0[/tex]
Tableau de signes
Racines : x + 2 = 0 ==> x = -2
-x - 2 = 0 ==> -x = 2
==> x = -2
[tex]\begin{array}{|c|ccccc|}x&-\infty&&-2&&+\infty\\x+2&&-&0&+& \\-x-2&&+&0&-& \\ (x+2)(-x-2)&&-&0&-\\ \end{array}\\\\\\S=\{-2\}[/tex]
