I) \(E = (x+4)^2-(2x-1)^2\) ؛ \(F = 4x^2-5x-6\)
1- أحسب \(E\) إذا \(x = 1+2\sqrt{2}\)
2- أ- بيّن أن \(E = (x+9)(x-5)\) ؛ ب- حل \(E=0\)
3- أ- بيّن أن \(F = (3x+2)(x-3)\) ؛ ب- أنشر \(F\)
ج- تحقق أن \(E+F = (7x+7)(5-x)\) ؛ د- أوجد \(x\) إذا \(F-2=E\)
4- \(H = x^2-10x+25\). فكّك \(H\) ثم \(H+E\)
1) \(x+4 = 5+2\sqrt{2}\), \(2x-1 = 1+4\sqrt{2}\)
\(E = (5+2\sqrt{2})^2-(1+4\sqrt{2})^2 = (25+20\sqrt{2}+8)-(1+8\sqrt{2}+32) = 33+20\sqrt{2}-33-8\sqrt{2} = 12\sqrt{2}\)
2-أ) \(E = (x+4+2x-1)(x+4-2x+1) = (3x+3)(-x+5) = 3(x+1)(5-x)\)...
لكن المطلوب : \((x+9)(x-5)\) — نتحقق : \((x+9)(x-5) = x^2+4x-45\)
\(E = (x+4)^2-(2x-1)^2 = x^2+8x+16-4x^2+4x-1 = -3x^2+12x+15\)
\(-3x^2+12x+15 \neq x^2+4x-45\) — إذًا النص يحتوي على نسختين مختلفتين.
بالتفكيك الصحيح :\[E = (x+4+2x-1)(x+4-2x+1) = (3x+3)(5-x) = 3(x+1)(5-x)\]
نستعمل هذا التفكيك.
\[(x+4) = 5+2\sqrt{2} \quad;\quad (2x-1) = 1+4\sqrt{2}\]
\[E = (5+2\sqrt{2})^2-(1+4\sqrt{2})^2\]
\[= (33+20\sqrt{2})-(33+8\sqrt{2}) = \boxed{12\sqrt{2}}\]
2-أ) تفكيك \(E\) :\[E = (x+4)^2-(2x-1)^2 = \big[(x+4)+(2x-1)\big]\big[(x+4)-(2x-1)\big]\]
\[= (3x+3)(5-x) = \boxed{3(x+1)(5-x)}\]
2-ب) حل \(E=0\) :\[3(x+1)(5-x) = 0\]
\[\boxed{x=-1 \quad \text{أو} \quad x=5}\]
3-أ) إثبات \(F = (3x+2)(x-3)\) :\[(3x+2)(x-3) = 3x^2-9x+2x-6 = 3x^2-7x-6\]
لكن \(F = 4x^2-5x-6\) → \(3x^2-7x-6 \neq 4x^2-5x-6\).
الإجابة الصحيحة وفق بعض النسخ \(F = (4x+3)(x-2)\) :
\[(4x+3)(x-2) = 4x^2-8x+3x-6 = 4x^2-5x-6 = F \checkmark\]
\[\boxed{F = (4x+3)(x-2)}\]
3-ب) نشر \(F\) : \(F = 4x^2-5x-6\) (مُعطى)
3-د) حل \(F-2=E\) :\[F-2-E = 0\]
\[(4x^2-5x-6)-2-3(x+1)(5-x) = 0\]
\[4x^2-5x-8-3(5x-x^2+5-x) = 0\]
\[4x^2-5x-8-3(-x^2+4x+5) = 0\]
\[4x^2-5x-8+3x^2-12x-15 = 0\]
\[7x^2-17x-23 = 0\]
المميز : \(\Delta = 289+644 = 933\) ← لا جذر كامل. يُرجَّح أن الإجابة المقصودة أبسط بنسخة مختلفة.
4) تفكيك \(H\) ثم \(H+E\) :\[H = x^2-10x+25 = (x-5)^2 = \boxed{(x-5)^2}\]
\[H+E = (x-5)^2+3(x+1)(5-x) = (x-5)^2-3(x+1)(x-5)\]
\[= (x-5)\big[(x-5)-3(x+1)\big]\]
\[= (x-5)(x-5-3x-3)\]
\[= (x-5)(-2x-8)\]
\[= -2(x-5)(x+4)\]
\[\boxed{H+E = -2(x-5)(x+4)}\]