I) \(a = (\sqrt{6}+2)(\sqrt{3}-\sqrt{2})\) ؛ \(b = \sqrt{3}^3-\sqrt{48}\)
1- بيّن أن \(a = \sqrt{2}\) و \(b = -\sqrt{3}\)
2- أحسب \((a+b)^2\)، \((a-b)^2\)، \((a-b)(a+b)\)
3- استنتج ضلع مربع مساحته \(S = 5-2\sqrt{6}\)
II) \(E = 4x^2+4x+1\) ؛ \(F = (3x+2)^2-(x+1)^2\)
1- فكّك \(E\) و \(F\)
2- أوجد \(x\) حيث \(E = F\)
I-1) \(a = \sqrt{18}-\sqrt{12}+2\sqrt{3}-2\sqrt{2} = 3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-2\sqrt{2} = \sqrt{2}\) ✓
\(b = 3\sqrt{3}-4\sqrt{3} = -\sqrt{3}\) ✓
I-2) \(a+b = \sqrt{2}-\sqrt{3}\) → \((a+b)^2 = 5-2\sqrt{6}\)
\(a-b = \sqrt{2}+\sqrt{3}\) → \((a-b)^2 = 5+2\sqrt{6}\)
\((a-b)(a+b) = a^2-b^2 = 2-3 = -1\)
I-3) \(S = 5-2\sqrt{6} = (a+b)^2 = (\sqrt{2}-\sqrt{3})^2\) → الضلع \(= |a+b| = |\sqrt{2}-\sqrt{3}| = \sqrt{3}-\sqrt{2}\)
II-1) \(E = (2x+1)^2\) ؛ \(F = (3x+2+x+1)(3x+2-x-1) = (4x+3)(2x+1)\)
II-2) \(E=F\) → \((2x+1)^2-(4x+3)(2x+1)=0\) → \((2x+1)(2x+1-4x-3)=0\) → \((2x+1)(-2x-2)=0\)
\[a = (\sqrt{6}+2)(\sqrt{3}-\sqrt{2}) = \sqrt{18}-\sqrt{12}+2\sqrt{3}-2\sqrt{2}\]
\[= 3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-2\sqrt{2} = (3-2)\sqrt{2}+(-2+2)\sqrt{3} = \sqrt{2} \checkmark\]
إثبات \(b = -\sqrt{3}\) :\[b = (\sqrt{3})^3-\sqrt{48} = 3\sqrt{3}-4\sqrt{3} = -\sqrt{3} \checkmark\]
I-2) الحسابات :\[(a+b)^2 = (\sqrt{2}-\sqrt{3})^2 = 2-2\sqrt{6}+3 = \boxed{5-2\sqrt{6}}\]
\[(a-b)^2 = (\sqrt{2}+\sqrt{3})^2 = 2+2\sqrt{6}+3 = \boxed{5+2\sqrt{6}}\]
\[(a-b)(a+b) = a^2-b^2 = (\sqrt{2})^2-(\sqrt{3})^2 = 2-3 = \boxed{-1}\]
I-3) ضلع المربع :\[S = 5-2\sqrt{6} = (a+b)^2 = (\sqrt{2}-\sqrt{3})^2\]
بما أن \(\sqrt{2} < \sqrt{3}\) → \(a+b = \sqrt{2}-\sqrt{3} < 0\)
→ الضلع \(= |a+b| = \sqrt{3}-\sqrt{2}\)
\[\boxed{c = \sqrt{3}-\sqrt{2}}\]
II-1) تفكيك \(E\) و \(F\) :\[E = 4x^2+4x+1 = (2x+1)^2 = \boxed{(2x+1)^2}\]
\[F = (3x+2)^2-(x+1)^2 = \big[(3x+2)+(x+1)\big]\big[(3x+2)-(x+1)\big] = (4x+3)(2x+1)\]
\[\boxed{F = (2x+1)(4x+3)}\]
II-2) حل \(E = F\) :\[(2x+1)^2-(2x+1)(4x+3) = 0\]
\[(2x+1)\big[(2x+1)-(4x+3)\big] = 0\]
\[(2x+1)(-2x-2) = 0\]
\[-2(2x+1)(x+1) = 0\]
\[\boxed{x = -\dfrac{1}{2} \quad \text{أو} \quad x = -1}\]