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Sagot :
Bonjour Chaton9070,
Exercice 1
[tex]A(x)=(-x+1)(2x-3)\\\\\begin{array}{|c|ccccccc|} x&-\infty&&\dfrac{3}{2}&&1&&+\infty\\&&&&&&&\\-x+1&&+&+&+&0&-&\\2x-3&&-&0&+&+&+&\\&&&&&&&\\A(x)&&-&0&+&0&-&\\ \end{array}\\\\\\\Longrightarrow\boxed{A(x)\ge0\Longleftrightarrow\ x\in[\dfrac{3}{2};1]}\\\text{ }\ \ \ \ \ \boxed{A(x)\le0\Longleftrightarrow\ x\in]-\infty;\dfrac{3}{2}]\cup[1;+\infty[}[/tex]
[tex]B(x)=\dfrac{-x-5}{3-2x}\\\\\begin{array}{|c|ccccccc|} x&-\infty&&-5&&\dfrac{3}{2}&&+\infty\\&&&&&&&\\-x-5&&+&0&-&-&-&\\3-2x&&+&+&+&0&-&\\&&&&&&&\\B(x)&&+&0&-&||&+&\\ \end{array}\\\\\\\Longrightarrow\boxed{B(x)\ge0\Longleftrightarrow\ x\in]-\infty;-5]\cup\ ]\dfrac{3}{2};+\infty[}\\\text{ }\ \ \ \ \ \boxed{B(x)\le0\Longleftrightarrow\ x\in[-5;\dfrac{3}{2}[}[/tex]
[tex]C(x)=\dfrac{1}{x-1}\\\\\begin{array}{|c|ccccc|} x&-\infty&&1&&+\infty\\&&&&&\\1&&+&+&+&\\x-1&&-&0&+&\\&&&&&\\C(x)&&-&||&+&\\ \end{array}\\\\\\\Longrightarrow\boxed{C(x)\ge0\Longleftrightarrow\ x\in\ ]1;+\infty[}\\\text{ }\ \ \ \ \ \boxed{C(x)\le0\Longleftrightarrow\ x\in\ ]-\infty;1[}[/tex]
Exercice 4
1) Développer f(x) et g(x)
[tex]f(x)=x(x+2)-(2x-1)(x+2)\\\\f(x)=x^2+2x-2x^2-4x+x+2\\\\\boxed{f(x)=-x^2-x+2}\\\\g(x)=(2x+3)^2-(x+1)^2\\\\g(x)=(4x^2+12x+9)-(x^2+2x+1)\\\\g(x)=4x^2+12x+9-x^2-2x-1\\\\\boxed{g(x)=3x^2+10x+8}[/tex]
2) Factoriser f(x) et g(x)
[tex]f(x)=x(x+2)-(2x-1)(x+2)\\\\f(x)=(x+2)[x-(2x-1)]\\\\f(x)=(x+2)[x-2x+1]\\\\\boxed{f(x)=(x+2)(-x+1)}\\\\\boxed{f(x)=(x+2)(1-x)}\\\\g(x)=(2x+3)^2-(x+1)^2\\\\g(x)=[(2x+3)-(x+1)][(2x+3)+(x+1)]\\\\g(x)=(2x+3-x-1)(2x+3+x+1)\\\\\boxed{g(x)=(x+2)(3x+4)}[/tex]
3) Calculs
[tex]f(x)=-x^2-x+2\\\\\Longrightarrow f(\sqrt{3})=-(\sqrt{3})^2-\sqrt{3}+2\\\\\Longrightarrow f(\sqrt{3})=-3-\sqrt{3}+2\\\\\Longrightarrow\boxed{f(\sqrt{3})=-1-\sqrt{3}}\\\\\\g(x)=3x^2+10x+8\\\\\Longrightarrow g(\sqrt{5})=3\times(\sqrt{5})^2+10\sqrt{5}+8\\\\\Longrightarrow g(\sqrt{5})=3\times5+10\sqrt{5}+8\\\\\Longrightarrow g(\sqrt{5})=15+10\sqrt{5}+8\\\\\Longrightarrow\boxed{g(\sqrt{5})=23+10\sqrt{5}}[/tex]
4) Equations.
[tex]f(x)=2\\\\-x^2-x+2=2\\\\-x^2-x=0\\\\-x(x+1)=0\\\\-x=0\ \ ou\ \ x+1=0\\\\\boxed{x=0\ \ ou\ \ x=-1}[/tex]
[tex]g(x)=0\\\\(x+2)(3x+4)=0\\\\x+2=0\ \ ou\ \ 3x+4=0\\\\\boxed{x=-2\ \ ou\ \ x=-\dfrac{4}{3}}[/tex]
[tex]f(x)=g(x)\\\\(x+2)(-x+1)=(x+2)(3x+4)\\\\(x+2)(-x+1)-(x+2)(3x+4)=0\\\\(x+2)[(-x+1)-(3x+4)]=0\\\\(x+2)(-x+1-3x-4)=0\\\\(x+2)(-4x-3)=0\\\\x+2=0\ \ ou\ \ -4x-3=0\\\\\boxed{x=-2\ \ ou\ \ x=-\dfrac{3}{4}}[/tex]
5) Inéquations
[tex]g(x) \ \textless \ 8\\\\3x^2+10x+8\ \textless \ 8\\\\3x^2+10x\ \textless \ 0\\\\x(3x+10)\ \textless \ 0\\\\\text{Tableau de signes...}\\\\\boxed{S=\ ]-\dfrac{10}{3};0[}\\\\\\f(x)\ge0\\\\(x+2)(-x+1)\ge0\\\\\text{Tableau de signes...}\\\\\boxed{S=]-\infty;-2]\cup[1;+\infty[}[/tex]
[tex]f(x)\ \textless \ g(x)\\\\(x+2)(-x+1)\ \textless \ (x+2)(3x+4)\\\\(x+2)(-x+1)-(x+2)(3x+4)\ \textless \ 0\\\\(x+2)[(-x+1)-(3x+4)]\ \textless \ 0\\\\(x+2)(-x+1-3x-4)\ \textless \ 0\\\\(x+2)(-4x-3)\ \textless \ 0\\\\\text{Tableau de signes...}\\\\\boxed{S=]-\infty;-2[\ \cup\ ]-\dfrac{3}{4};+\infty[}[/tex]
Exercice 1
[tex]A(x)=(-x+1)(2x-3)\\\\\begin{array}{|c|ccccccc|} x&-\infty&&\dfrac{3}{2}&&1&&+\infty\\&&&&&&&\\-x+1&&+&+&+&0&-&\\2x-3&&-&0&+&+&+&\\&&&&&&&\\A(x)&&-&0&+&0&-&\\ \end{array}\\\\\\\Longrightarrow\boxed{A(x)\ge0\Longleftrightarrow\ x\in[\dfrac{3}{2};1]}\\\text{ }\ \ \ \ \ \boxed{A(x)\le0\Longleftrightarrow\ x\in]-\infty;\dfrac{3}{2}]\cup[1;+\infty[}[/tex]
[tex]B(x)=\dfrac{-x-5}{3-2x}\\\\\begin{array}{|c|ccccccc|} x&-\infty&&-5&&\dfrac{3}{2}&&+\infty\\&&&&&&&\\-x-5&&+&0&-&-&-&\\3-2x&&+&+&+&0&-&\\&&&&&&&\\B(x)&&+&0&-&||&+&\\ \end{array}\\\\\\\Longrightarrow\boxed{B(x)\ge0\Longleftrightarrow\ x\in]-\infty;-5]\cup\ ]\dfrac{3}{2};+\infty[}\\\text{ }\ \ \ \ \ \boxed{B(x)\le0\Longleftrightarrow\ x\in[-5;\dfrac{3}{2}[}[/tex]
[tex]C(x)=\dfrac{1}{x-1}\\\\\begin{array}{|c|ccccc|} x&-\infty&&1&&+\infty\\&&&&&\\1&&+&+&+&\\x-1&&-&0&+&\\&&&&&\\C(x)&&-&||&+&\\ \end{array}\\\\\\\Longrightarrow\boxed{C(x)\ge0\Longleftrightarrow\ x\in\ ]1;+\infty[}\\\text{ }\ \ \ \ \ \boxed{C(x)\le0\Longleftrightarrow\ x\in\ ]-\infty;1[}[/tex]
Exercice 4
1) Développer f(x) et g(x)
[tex]f(x)=x(x+2)-(2x-1)(x+2)\\\\f(x)=x^2+2x-2x^2-4x+x+2\\\\\boxed{f(x)=-x^2-x+2}\\\\g(x)=(2x+3)^2-(x+1)^2\\\\g(x)=(4x^2+12x+9)-(x^2+2x+1)\\\\g(x)=4x^2+12x+9-x^2-2x-1\\\\\boxed{g(x)=3x^2+10x+8}[/tex]
2) Factoriser f(x) et g(x)
[tex]f(x)=x(x+2)-(2x-1)(x+2)\\\\f(x)=(x+2)[x-(2x-1)]\\\\f(x)=(x+2)[x-2x+1]\\\\\boxed{f(x)=(x+2)(-x+1)}\\\\\boxed{f(x)=(x+2)(1-x)}\\\\g(x)=(2x+3)^2-(x+1)^2\\\\g(x)=[(2x+3)-(x+1)][(2x+3)+(x+1)]\\\\g(x)=(2x+3-x-1)(2x+3+x+1)\\\\\boxed{g(x)=(x+2)(3x+4)}[/tex]
3) Calculs
[tex]f(x)=-x^2-x+2\\\\\Longrightarrow f(\sqrt{3})=-(\sqrt{3})^2-\sqrt{3}+2\\\\\Longrightarrow f(\sqrt{3})=-3-\sqrt{3}+2\\\\\Longrightarrow\boxed{f(\sqrt{3})=-1-\sqrt{3}}\\\\\\g(x)=3x^2+10x+8\\\\\Longrightarrow g(\sqrt{5})=3\times(\sqrt{5})^2+10\sqrt{5}+8\\\\\Longrightarrow g(\sqrt{5})=3\times5+10\sqrt{5}+8\\\\\Longrightarrow g(\sqrt{5})=15+10\sqrt{5}+8\\\\\Longrightarrow\boxed{g(\sqrt{5})=23+10\sqrt{5}}[/tex]
4) Equations.
[tex]f(x)=2\\\\-x^2-x+2=2\\\\-x^2-x=0\\\\-x(x+1)=0\\\\-x=0\ \ ou\ \ x+1=0\\\\\boxed{x=0\ \ ou\ \ x=-1}[/tex]
[tex]g(x)=0\\\\(x+2)(3x+4)=0\\\\x+2=0\ \ ou\ \ 3x+4=0\\\\\boxed{x=-2\ \ ou\ \ x=-\dfrac{4}{3}}[/tex]
[tex]f(x)=g(x)\\\\(x+2)(-x+1)=(x+2)(3x+4)\\\\(x+2)(-x+1)-(x+2)(3x+4)=0\\\\(x+2)[(-x+1)-(3x+4)]=0\\\\(x+2)(-x+1-3x-4)=0\\\\(x+2)(-4x-3)=0\\\\x+2=0\ \ ou\ \ -4x-3=0\\\\\boxed{x=-2\ \ ou\ \ x=-\dfrac{3}{4}}[/tex]
5) Inéquations
[tex]g(x) \ \textless \ 8\\\\3x^2+10x+8\ \textless \ 8\\\\3x^2+10x\ \textless \ 0\\\\x(3x+10)\ \textless \ 0\\\\\text{Tableau de signes...}\\\\\boxed{S=\ ]-\dfrac{10}{3};0[}\\\\\\f(x)\ge0\\\\(x+2)(-x+1)\ge0\\\\\text{Tableau de signes...}\\\\\boxed{S=]-\infty;-2]\cup[1;+\infty[}[/tex]
[tex]f(x)\ \textless \ g(x)\\\\(x+2)(-x+1)\ \textless \ (x+2)(3x+4)\\\\(x+2)(-x+1)-(x+2)(3x+4)\ \textless \ 0\\\\(x+2)[(-x+1)-(3x+4)]\ \textless \ 0\\\\(x+2)(-x+1-3x-4)\ \textless \ 0\\\\(x+2)(-4x-3)\ \textless \ 0\\\\\text{Tableau de signes...}\\\\\boxed{S=]-\infty;-2[\ \cup\ ]-\dfrac{3}{4};+\infty[}[/tex]
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