Correction - S'auto-évaluer Recopier et compléter page 179 - Activités algébriques
Recopions et complétons.
1- Si \(x = 0\) alors \((x + 1)^2 – (2x + 1)^2 = ...\)
\((x + 1)^2 – (2x + 1)^2 = (x^2+2.x.1+1^2)-((2x)^2+2.2x.1+1^2)\) \(\Rightarrow \)
\((x + 1)^2 – (2x + 1)^2 = (x^2+2x+1)-(4x^2+4x+1)\) \(\Rightarrow \)
\((x + 1)^2 – (2x + 1)^2 = x^2+2x+1-4x^2-4x-1\) \(\Rightarrow \)
\((x + 1)^2 – (2x + 1)^2 = -3x^2-2x=x(3x-2)\)
Donc si \(x = 0\) alors \(x(3x-2)=0(3\times 0-2)=0\times -2=0\).
Conclusion :
Si \(x = 0\) alors \((x + 1)^2 – (2x + 1)^2 = 0\)
2- \((2x – 1)(4x^2 + 4x + 1)\) est une factorisation de \(...\)
\((2x – 1)(4x^2 + 4x + 1)=8x^3+8x^2+2x-4x^2-4x-1\) \(\Rightarrow \)
\((2x – 1)(4x^2 + 4x + 1)=8x^3+4x^2-2x-1\)
3- \(1,61^2 + 2 \times1,61\times0,39 + 0,39^2 = ...\)
\((a+b)^2=a^2+2.a.b+b^2\) \(\Leftrightarrow\) \((a+b)^2=a^2+2.a.b+b^2\)
\(1,61^2+2\times1,61\times0,39+0,39^2=(1,61+0,39) ^2=2^2=4\)
1- Si \(x = 0\) alors \((x + 1)^2 – (2x + 1)^2 = ...\)
\((x + 1)^2 – (2x + 1)^2 = (x^2+2.x.1+1^2)-((2x)^2+2.2x.1+1^2)\) \(\Rightarrow \)
\((x + 1)^2 – (2x + 1)^2 = (x^2+2x+1)-(4x^2+4x+1)\) \(\Rightarrow \)
\((x + 1)^2 – (2x + 1)^2 = x^2+2x+1-4x^2-4x-1\) \(\Rightarrow \)
\((x + 1)^2 – (2x + 1)^2 = -3x^2-2x=x(3x-2)\)
Donc si \(x = 0\) alors \(x(3x-2)=0(3\times 0-2)=0\times -2=0\).
Conclusion :
Si \(x = 0\) alors \((x + 1)^2 – (2x + 1)^2 = 0\)
2- \((2x – 1)(4x^2 + 4x + 1)\) est une factorisation de \(...\)
\((2x – 1)(4x^2 + 4x + 1)=8x^3+8x^2+2x-4x^2-4x-1\) \(\Rightarrow \)
\((2x – 1)(4x^2 + 4x + 1)=8x^3+4x^2-2x-1\)
Conclusion :
\((2x – 1)(4x^2 + 4x + 1)\) est une factorisation de \(8x^3+4x^2-2x-1\)3- \(1,61^2 + 2 \times1,61\times0,39 + 0,39^2 = ...\)
\((a+b)^2=a^2+2.a.b+b^2\) \(\Leftrightarrow\) \((a+b)^2=a^2+2.a.b+b^2\)
\(1,61^2+2\times1,61\times0,39+0,39^2=(1,61+0,39) ^2=2^2=4\)
Conclusion :
\(1,61^2 + 2 \times1,61\times0,39 + 0,39^2 =4\).
Libellés:
1ère année secondaire
Activités algébriques
Correction
Corrigées
exercice
Le Mathématicien
manuel scolaire
Math
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