\[E = 2x^2 + 6\sqrt{2}x + 5 \qquad F = (\sqrt{2}x - 3)(\sqrt{2}x + 1)\]
1- أ- أحسب \(E\) إذا كان \(x = -\sqrt{2}\)
ب- أحسب \(F\) حيث \(x = -1\)
2- أ- بيّن أن \(E = (\sqrt{2}x + 3)^2 - 4\)
ب- استنتج تفكيكاً للعبارة \(E\)
3- بيّن أن \(E + F = 2(\sqrt{2}x + 1)^2\)
4- أ- أوجد \(x\) حيث \(2x^2 + 6\sqrt{2}x + 5 = (\sqrt{2}x-3)(\sqrt{2}x+1)\)
ب- أوجد \(x\) حيث \(\sqrt{E+F} = 2\sqrt{2}\)
1-أ) \(E = 2(-\sqrt{2})^2 + 6\sqrt{2}(-\sqrt{2}) + 5 = 4 - 12 + 5 = -3\)
1-ب) \(F = (\sqrt{2}(-1)-3)(\sqrt{2}(-1)+1) = (-\sqrt{2}-3)(1-\sqrt{2})\)
2-أ) نتحقق بالنشر : \((\sqrt{2}x+3)^2 - 4 = 2x^2+6\sqrt{2}x+9-4 = 2x^2+6\sqrt{2}x+5 = E\)
2-ب) \(E = (\sqrt{2}x+3+2)(\sqrt{2}x+3-2) = (\sqrt{2}x+5)(\sqrt{2}x+1)\)
3) \(F = (\sqrt{2}x)^2 + \sqrt{2}x - 3\sqrt{2}x - 3 = 2x^2-2\sqrt{2}x-3\)
→ \(E+F = (2x^2+6\sqrt{2}x+5)+(2x^2-2\sqrt{2}x-3) = 4x^2+4\sqrt{2}x+2 = 2(2x^2+2\sqrt{2}x+1) = 2(\sqrt{2}x+1)^2\)
4-أ) \(E = F\) → \(E-F=0\) → utiliser les factorisations
4-ب) \(E+F = 8\) → \(2(\sqrt{2}x+1)^2 = 8\) → \((\sqrt{2}x+1)^2 = 4\)
\[E = 2(-\sqrt{2})^2 + 6\sqrt{2} \times (-\sqrt{2}) + 5 = 2 \times 2 - 6 \times 2 + 5 = 4 - 12 + 5 = \boxed{-3}\]
1-ب) حساب \(F\) عند \(x = -1\) :\[F = (\sqrt{2} \times (-1) - 3)(\sqrt{2} \times (-1) + 1) = (-\sqrt{2}-3)(1-\sqrt{2})\]
\[= -\sqrt{2}+2-3+3\sqrt{2} = 2\sqrt{2}-1\]
\[\boxed{F = 2\sqrt{2}-1}\]
2-أ) إثبات \(E = (\sqrt{2}x+3)^2 - 4\) :\[(\sqrt{2}x+3)^2 - 4 = 2x^2 + 6\sqrt{2}x + 9 - 4 = 2x^2 + 6\sqrt{2}x + 5 = E \checkmark\]
2-ب) تفكيك \(E\) :\[E = (\sqrt{2}x+3)^2 - 2^2 = \big[(\sqrt{2}x+3)+2\big]\big[(\sqrt{2}x+3)-2\big]\]
\[\boxed{E = (\sqrt{2}x+5)(\sqrt{2}x+1)}\]
3) إثبات \(E + F = 2(\sqrt{2}x+1)^2\) :ننشر \(F\) أولًا :
\[F = (\sqrt{2}x-3)(\sqrt{2}x+1) = 2x^2 + \sqrt{2}x - 3\sqrt{2}x - 3 = 2x^2 - 2\sqrt{2}x - 3\]
\[E + F = (2x^2+6\sqrt{2}x+5) + (2x^2-2\sqrt{2}x-3) = 4x^2 + 4\sqrt{2}x + 2\]
\[= 2(2x^2 + 2\sqrt{2}x + 1) = 2(\sqrt{2}x+1)^2 \checkmark\]
4-أ) حل \(E = F\) :\[E - F = 0\]
باستعمال التفكيكين :
\[(\sqrt{2}x+5)(\sqrt{2}x+1) - (\sqrt{2}x-3)(\sqrt{2}x+1) = 0\]
\[(\sqrt{2}x+1)\big[(\sqrt{2}x+5)-(\sqrt{2}x-3)\big] = 0\]
\[(\sqrt{2}x+1)(8) = 0\]
\[\sqrt{2}x + 1 = 0\]
\[\boxed{x = -\dfrac{1}{\sqrt{2}} = -\dfrac{\sqrt{2}}{2}}\]
4-ب) حل \(\sqrt{E+F} = 2\sqrt{2}\) :\[\sqrt{E+F} = 2\sqrt{2} \Rightarrow E+F = 8\]
من 3) : \(2(\sqrt{2}x+1)^2 = 8 \Rightarrow (\sqrt{2}x+1)^2 = 4\)
\[\sqrt{2}x+1 = 2 \quad \text{أو} \quad \sqrt{2}x+1 = -2\]
\[\sqrt{2}x = 1 \quad \text{أو} \quad \sqrt{2}x = -3\]
\[\boxed{x = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \quad \text{أو} \quad x = \dfrac{-3}{\sqrt{2}} = \dfrac{-3\sqrt{2}}{2}}\]