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Sagot :
[tex]f(x)+g(x)=x^3+2x^2+3x^2-1=x^3+5x^2-1\\
D:x\in\mathbb{R}[/tex]
[tex]f(x)-g(x)=x^3+2x^2-(3x^2-1)=x^3+2x^2-3x^2+1=x^3-x^2+1\\ D:x\in\mathbb{R}[/tex]
[tex]f(x)\cdotg(x)=(x^3+2x^2)(3x^2-1)=3x^5-x^3+6x^4-2x^2=\\ 3x^5+6x^4-x^3-2x^2\\ D:x\in\mathbb{R}[/tex]
[tex]\frac{f(x)}{g(x)}=\frac{x^3+2x^2}{3x^2-1}\\ 3x^2-1\not=0\\ 3x^2\not=1\\ x^2\not=\frac{1}{3}\\ x\not =-\frac{\sqrt3}{3} \wedge x\not =\frac{\sqrt3}{3}\\ D:x\in(-\infty,-\frac{\sqrt3}{3} )\cup(\frac{\sqrt3}{3} ,\infty) [/tex]
[tex]f(x)-g(x)=x^3+2x^2-(3x^2-1)=x^3+2x^2-3x^2+1=x^3-x^2+1\\ D:x\in\mathbb{R}[/tex]
[tex]f(x)\cdotg(x)=(x^3+2x^2)(3x^2-1)=3x^5-x^3+6x^4-2x^2=\\ 3x^5+6x^4-x^3-2x^2\\ D:x\in\mathbb{R}[/tex]
[tex]\frac{f(x)}{g(x)}=\frac{x^3+2x^2}{3x^2-1}\\ 3x^2-1\not=0\\ 3x^2\not=1\\ x^2\not=\frac{1}{3}\\ x\not =-\frac{\sqrt3}{3} \wedge x\not =\frac{\sqrt3}{3}\\ D:x\in(-\infty,-\frac{\sqrt3}{3} )\cup(\frac{\sqrt3}{3} ,\infty) [/tex]
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