Correction - Exercice 06 page 180 - Activités algébriques
1ère année secondaire
Activités algébriques
Exercice 06 page 180
Développons et réduisons les expressions suivantes
a) \((\sqrt{2}–\frac{1}{2})(2+\frac{\sqrt{2}}{2}+\frac{1}{4} )\).
\((\sqrt{2}–\frac{1}{2})(2+\frac{\sqrt{2}}{2}+\frac{1}{4} )\)
\(=2\sqrt{2}+\frac{\sqrt{2}\times\sqrt{2}}{2}+\frac{\sqrt{2}\times1}{4}-\frac{2}{2}-\frac{1\times\sqrt{2}}{2\times2}-\frac{1\times1}{2\times4}\)
\(=2\sqrt{2}+\frac{2}{2}+\frac{\sqrt{2}}{4}-\frac{2}{2}-\frac{\sqrt{2}}{4}-\frac{1}{8}\)
\(=2\sqrt{2}-\)\(\frac{1}{8}\)
b) \((\frac{1}{3}a–\frac{2}{5}b)(\frac{1}{3}a+\frac{2}{5}b)\).
\((\frac{1}{3}a–\frac{2}{5}b)(\frac{1}{3}a+\frac{2}{5}b)\)
\(=(\frac{1}{3}a)^2–(\frac{2}{5}b)^2\)
\(=\frac{1}{9}a^2–\frac{4}{25}b^2\)
c) \((2a–1)^2-(2a+1)(3a+5)\).
\((2a–1)^2-(2a+1)(3a+5)\)
\(=2.2a^2-2.2a.1+1^2-(2a\times3a+2a\times5+1\times3a+1\times5)\)
\(=4a^2-4a+1-(6a^2+10a+3a+5)\)
\(=4a^2-4a+1-(6a^2+13a+5)\)
\(=4a^2-4a+1-6a^2-13a-5)\)
\(=-2a^2-17a-4\)
d) \((2x–1)^3+3(2x–1)^2(1-x)+3(2x–1)(1–x)^2+(1–x)^3\).
\((2x–1)^3+3(2x–1)^2(1-x)+3(2x–1)(1–x)^2+(1–x)^3\)
\(=((2x-1)+(1-x))^3\)
a) \((\sqrt{2}–\frac{1}{2})(2+\frac{\sqrt{2}}{2}+\frac{1}{4} )\).
\((\sqrt{2}–\frac{1}{2})(2+\frac{\sqrt{2}}{2}+\frac{1}{4} )\)
\(=2\sqrt{2}+\frac{\sqrt{2}\times\sqrt{2}}{2}+\frac{\sqrt{2}\times1}{4}-\frac{2}{2}-\frac{1\times\sqrt{2}}{2\times2}-\frac{1\times1}{2\times4}\)
\(=2\sqrt{2}+\frac{2}{2}+\frac{\sqrt{2}}{4}-\frac{2}{2}-\frac{\sqrt{2}}{4}-\frac{1}{8}\)
\(=2\sqrt{2}-\)\(\frac{1}{8}\)
b) \((\frac{1}{3}a–\frac{2}{5}b)(\frac{1}{3}a+\frac{2}{5}b)\).
\((\frac{1}{3}a–\frac{2}{5}b)(\frac{1}{3}a+\frac{2}{5}b)\)
\(=(\frac{1}{3}a)^2–(\frac{2}{5}b)^2\)
\(=\frac{1}{9}a^2–\frac{4}{25}b^2\)
c) \((2a–1)^2-(2a+1)(3a+5)\).
\((2a–1)^2-(2a+1)(3a+5)\)
\(=2.2a^2-2.2a.1+1^2-(2a\times3a+2a\times5+1\times3a+1\times5)\)
\(=4a^2-4a+1-(6a^2+10a+3a+5)\)
\(=4a^2-4a+1-(6a^2+13a+5)\)
\(=4a^2-4a+1-6a^2-13a-5)\)
\(=-2a^2-17a-4\)
d) \((2x–1)^3+3(2x–1)^2(1-x)+3(2x–1)(1–x)^2+(1–x)^3\).
\((2x–1)^3+3(2x–1)^2(1-x)+3(2x–1)(1–x)^2+(1–x)^3\)
\(=((2x-1)+(1-x))^3\)
\(=(2x-1+1-x)^3\)
\(=x^3\)
e) \((2-\frac{3}{4}t)^2+(\frac{1}{4}t+1)^2\).
\((2-\frac{3}{4}t)^2+(\frac{1}{4}t+1)^2\)
\(=2^2-2\times2\times\frac{3}{4}t+(\frac{3}{4}t)^2+(\frac{1}{4}t)^2+2\times\frac{1}{4}t\times1+1^2\)
\(=4-\frac{12}{4}t+\frac{9}{16}t^2+\frac{1}{16}t^2+\frac{2}{4}t+1\)
\(=\frac{10}{16}t^2-\frac{10}{4}t+5\)
\(=\frac{5}{8}t^2-\frac{5}{2}t+5\)
f) \((x–2)(x+3)(x^2-1)–(x^3+2x)(x–6)\).
\((x–2)(x+3)(x^2-1)–(x^3+2x)(x–6)\)
\(=(x^2+3x-2x-6)(x^2-1)-(x^4-6x^3+2x^2-12x)\)
\(=(x^2+x-6)(x^2-1)-x^4+6x^3-2x^2+12x)\)
\(=x^4-x^2+x^3-x-6x^2+6-x^4+6x^3-2x^2+12x\)
\(=7x^3-9x^2+11x+6\)
g) \((π+x)^3–3π(π +x)+2π^3–x^3\).
\((π+x)^3–3π(π +x)+2π^3–x^3\)
\(=\pi^3+3\pi^2x+3\pi x^2+x^3-3\pi^2-3\pi x+2\pi^3-x^3\)
\(=3\pi^3+3\pi^2x+3\pi x^2-3\pi^2-3\pi x\)
Libellés:
1ère année secondaire
Activités algébriques
Correction
Corrigées
exercice
Le Mathématicien
manuel scolaire
Math
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