Correction - Exercice 07 page 180 - Activités algébriques
1ère année secondaire
Activités algébriques
Exercice 07 page 180
Factorisons les expressions suivantes :
a) \(\frac{x^2}{25}\)\(-\)\(\frac{y^2}{16}\).
\(\frac{x^2}{25}\)\(-\)\(\frac{y^2}{16}\)\(=\)
\((\frac{x}{5})^2\)\(-\)\((\frac{y}{4})^2\)\(=\)
\((\frac{x}{5}\)\(-\)\(\frac{y}{4})\)\((\frac{x}{5}\)\(+\)\(\frac{y}{4})\)\(a^2-b^2=(a-b)(a+b)\)
b) \((7\sqrt{2}x)^2-14\sqrt{2}.x.y+y^2\).
a) \(\frac{x^2}{25}\)\(-\)\(\frac{y^2}{16}\).
\(\frac{x^2}{25}\)\(-\)\(\frac{y^2}{16}\)\(=\)
\((\frac{x}{5})^2\)\(-\)\((\frac{y}{4})^2\)\(=\)
\((\frac{x}{5}\)\(-\)\(\frac{y}{4})\)\((\frac{x}{5}\)\(+\)\(\frac{y}{4})\)\(a^2-b^2=(a-b)(a+b)\)
b) \((7\sqrt{2}x)^2-14\sqrt{2}.x.y+y^2\).
\((7\sqrt{2}x)^2-14\sqrt{2}.x.y+y^2\)\(=\)
\((7\sqrt{2}x-y)^2\)
\((a-b)^2=a^2-2.a.b+b^2\)
c) \((2t–1)^3+8\).
\((2t–1)^3+8\)\(=\)\((2t–1)^3+2^3\)\(=\)
\((2t-1+2)((2t–1)^2-2(2t-1)+2^2)\)\(=\)
\((2t+1)((4t^2-2\times2t\times1+1)-2(2t)+4)\)\(=\)
\((2t+1)(4t^2-4t+1-4t+2+4)\)\(=\)
\((2t+1)(4t^2-8t+7)\)
\(a^3+b^3=(a+b)(a^2-ab+b^2)\)
d) \((\frac{3}{2}k+\frac{1}{2})^3+64\).
\((\frac{3}{2}k+\frac{1}{2})^3+64\)\(=\)\((\frac{3}{2}k+\frac{1}{2})^3+4^3\)\(=\)
\(((\frac{3}{2}k+\frac{1}{2})+4)((\frac{3}{2}k+\frac{1}{2})^2-4(\frac{3}{2}k+\frac{1}{2})+4^2)\)\(=\)
d) \((\frac{3}{2}k+\frac{1}{2})^3+64\).
\((\frac{3}{2}k+\frac{1}{2})^3+64\)\(=\)\((\frac{3}{2}k+\frac{1}{2})^3+4^3\)\(=\)
\(((\frac{3}{2}k+\frac{1}{2})+4)((\frac{3}{2}k+\frac{1}{2})^2-4(\frac{3}{2}k+\frac{1}{2})+4^2)\)\(=\)
\((\frac{3}{2}k+\frac{9}{2})((\frac{9}{4}k^2+\frac{6}{4}k+\frac{1}{4})-\frac{12}{2}k+\frac{4}{2}+16)\)\(=\)
\((\frac{3}{2}k+\frac{9}{2})(\frac{9}{4}k^2-\frac{9}{2}k-\frac{7}{4}+16)\)\(=\)
\((\frac{3}{2}k+\frac{9}{2})(\frac{9}{4}k^2-\frac{9}{2}k+\frac{57}{4})\)
\((\frac{3}{2}k+\frac{9}{2})(\frac{9}{4}k^2-\frac{9}{2}k+\frac{57}{4})\)
\(a^3+b^3=(a+b)(a^2-ab+b^2)\)
e) \(x^4–x^2\)\(+\)\(\frac{1}{4}\).
\(x^4–x^2\)\(+\)\(\frac{1}{4}=\)
\((x^2)^2-2\times\frac{1}{2}\times x^2+(\frac{1}{2})^2\)\(=\)
e) \(x^4–x^2\)\(+\)\(\frac{1}{4}\).
\(x^4–x^2\)\(+\)\(\frac{1}{4}=\)
\((x^2)^2-2\times\frac{1}{2}\times x^2+(\frac{1}{2})^2\)\(=\)
\((x^2-\frac{1}{2})^2\)
\(a^2-2.a.b+b^2=(a-b)^2\)
f) \(27\times10^{-3} + 27\times10^{-4} + 9\times10^{-5} + 10^{-6}\).
\(1728x^3-\frac{8}{125}=\)
\((12x)^3-(\frac{2}{5})^3=\)
\((12x-\frac{2}{5})((12x)^2+12x\times\frac{2}{5}+(\frac{2}{5})^2)=\)
\((12x-\frac{2}{5})(144x^2+\frac{24}{5}x+\frac{4}{25})\)
h) \(\pi^3+3\pi^2+3\pi+1\).
f) \(27\times10^{-3} + 27\times10^{-4} + 9\times10^{-5} + 10^{-6}\).
\(27\times10^{-3}+27\times10^{-4}+9\times10^{-5}+10^{-6}=\)
\((3\times10^{-1})^{3}+3\times(3\times10^{-1})^2\times10^{-2}+3\times(3\times10^{-1})\times(10^{-2})^2+(10^{-2})^{3}=\)
\(((3\times10^{-1})+10^{-2})^3\)
\(a^3+3.a^2.b+3.a.b^2+b^3=(a+b)^3\)
g) \(1728x^3–\frac{8}{125}\).
\((3\times10^{-1})^{3}+3\times(3\times10^{-1})^2\times10^{-2}+3\times(3\times10^{-1})\times(10^{-2})^2+(10^{-2})^{3}=\)
\(((3\times10^{-1})+10^{-2})^3\)
\(a^3+3.a^2.b+3.a.b^2+b^3=(a+b)^3\)
g) \(1728x^3–\frac{8}{125}\).
\(1728x^3-\frac{8}{125}=\)
\((12x)^3-(\frac{2}{5})^3=\)
\((12x-\frac{2}{5})((12x)^2+12x\times\frac{2}{5}+(\frac{2}{5})^2)=\)
\((12x-\frac{2}{5})(144x^2+\frac{24}{5}x+\frac{4}{25})\)
h) \(\pi^3+3\pi^2+3\pi+1\).
\(\pi^3+3\pi^2+3\pi+1=\)
\(\pi^3+3.\pi^2.1+3.\pi.1^2+1^3=\)
\((\pi+1)^3\)
Libellés:
1ère année secondaire
Activités algébriques
Correction
Corrigées
exercice
Le Mathématicien
manuel scolaire
Math
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