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L'évaluation du polynôme \(P(x)= -3x^2+x-4\) en \(x=\frac{1}{2}\) vaut
\(-5\)
\(-\frac{17}{4}\)
\(-\frac{5}{2}\)
\(-\frac{9}{2}\)
Effectuez \((-4x^2+2y^3)^2\)
\(16x^4+4y^5-16x^2y^3\)
\(16x^4+4y^6-16x^2y^3\)
\(4y^6-16x^4\)
\(4x^4+2y^6-8x^2y^3\)
Factorisez \(x^8+y^8+x^4y^4\)
\((x^4+y^4-x^2y^2)(x^4+y^4+x^2y^2)\)
\((x^2-y^2)^2(x^2+y^2)^2\)
\(x^4(x^4+y^4)+y^8\)
impossible
Factorisez \(ac+bc+ad+bd\)
\((a+d)(b+c)\)
\(a(b+c+d)\)
\((a+b)(c+d)\)
Factorisez \((a+b)^3-(a+b)\)
\((a+b)(a^2+2ab+b^2)\)
\((a+b)^2\)
\((a+b)(a^2+2ab+b^2-1)\)
\(a^3+b^3-a-b\)
\((x^2-1)^3=\)
\(x^6-1\)
\(-x^6+3x^4-3x^2+1\)
\(x^6-3x^4+3x^2-1\)
\(x^5-3x^4+3x^2-1\)
Effectuez \((x+\frac{1}{x})^3\)
\(\dfrac{x^9+3x^4+3x^2+1}{x^3}\)
\(\dfrac{x^6+3x^4+3x^2+1}{x^3}\)
\(\dfrac{x^6+3x^5+3x+1}{x^3}\)
\(\dfrac{x^6+1}{x^3}\)
Factorisez \(x^7-3x^5+3x^3-x\)
\(x(x-1)^3(x+1)^3\)
\(x(x^2-1)(x^4-3x^2-1)\)
\(x(x^2-1)(x^4-3x^3+x^2+1)\)
Factorisez \(3(2-x)^2-3(x-2)^3\)
\(3(2-x)^2(7-3x)\)
\(3-x\)
\(3(2-x)^2(3-x)\)
\(-1-x\)
Factorisez \(x^3-5x^2+5x-1=\)
\((x-1)^5\)
\((x-1)(x^2-6x+1)\)
\((x-1)(x^2-4x+1)\)
\((x-1)^3\)
