Bonsoir Morgane98bla
[tex]d)\ (x+1)^2\ \boxed{}\ (2x-3)^2\\\\(x+1)^2-(2x-3)^2\ \boxed{}\ 0\\\\\ [(x+1)+(2x-3)][(x+1)-(2x-3)]\ \boxed{}\ 0\\\\(x+1+2x-3)(x+1-2x+3)\ \boxed{}\ 0\\\\(3x-2)(-x+4)\ \boxed{}\ 0[/tex]
Tableau de signes de (3x-2)(-x+4)
Racines : 2/3 ; 4
[tex]\begin{array}{|c|ccccccc||}x&-\infty&&\dfrac{2}{3}&&4&&+\infty\\ 3x-2&&-&0&+&+&+&\\ -x+4&&+&+&+&0&-&\\ (3x-2)(-x+4)&&-&0&+&0&-&\\\end{array}[/tex]
Par conséquent,
[tex] (x+1)^2< (2x-3)^2\Longleftrightarrow (3x-2)(-x+4)<0\\\\ (x+1)^2< (2x-3)^2\Longleftrightarrow x\in ]-\infty;\dfrac{2}{3}[\ \cup\ ]4;+\infty[\\\\ \boxed{(x+1)^2< (2x-3)^2\Longleftrightarrow S=]-\infty;\dfrac{2}{3}[\ \cup\ ]4;+\infty[}[/tex]
[tex](x+1)^2> (2x-3)^2\Longleftrightarrow (3x-2)(-x+4)>0\\\\ (x+1)^2> (2x-3)^2\Longleftrightarrow x\in ]\dfrac{2}{3}; 4[\\\\ \boxed{(x+1)^2> (2x-3)^2\Longleftrightarrow S=]\dfrac{2}{3}; 4[}[/tex]
[tex]f)\ \dfrac{4x-7}{3x+2}\ \boxed{}\ 4\\\\\dfrac{4x-7}{3x+2}-4\ \boxed{}\ 0\\\\\dfrac{4x-7-4(3x+2)}{3x+2}\ \boxed{}\ 0\\\\\dfrac{4x-7-12x-8}{3x+2}\ \boxed{}\ 0\\\\\dfrac{-8x-15}{3x+2}\ \boxed{}\ 0[/tex]
Tableau de signes de (-8x-15)/(3x+2)
Racines : numérateur : -15/8
dénominateur : -2/3
[tex]\begin{array}{|c|ccccccc||}x&-\infty&&-\dfrac{15}{8}&&-\dfrac{2}{3}&&+\infty\\ -8x-15&&+&0&-&-&-&\\ 3x+2&&-&-&-&0&+&\\ (3x-2)(-x+4)&&-&0&+&||&-&\\\end{array}[/tex]
[tex] \dfrac{4x-7}{3x+2}<4\Longleftrightarrow \dfrac{-8x-15}{3x+2}<0\\\\ \dfrac{4x-7}{3x+2}<4\Longleftrightarrow x\in ]-\infty;-\dfrac{15}{8}[\ \cup\ ]-\dfrac{2}{3};+\infty[\\\\ \boxed{ \dfrac{4x-7}{3x+2}<4\Longleftrightarrow S=]-\infty;;-\dfrac{15}{8}[\ \cup\ ]-\dfrac{2}{3};+\infty[}[/tex]
[tex]\dfrac{4x-7}{3x+2}>4\Longleftrightarrow \dfrac{-8x-15}{3x+2}>0\\\\ \dfrac{4x-7}{3x+2}>4\Longleftrightarrow x\in ]-\dfrac{15}{8};-\dfrac{2}{3}[\\\\ \boxed{ \dfrac{4x-7}{3x+2}>4\Longleftrightarrow S=]-\dfrac{15}{8};-\dfrac{2}{3}[}[/tex]
