الجذاآت المعتبرة — التمرين 10

✅ التصحيح المفصّل

1) حساب \(B\) عند \(x = 2\) :

\[B = (2\times2+1)^2-2(2\times2+5) = 25-18 = \boxed{7}\]

2-أ) نشر واختصار \(B\) :

\[B = (2x+1)^2-2(2x+5) = 4x^2+4x+1-4x-10 = \boxed{4x^2-9}\]

2-ب) تفكيك \(B\) :

\[B = (2x)^2-3^2 = \boxed{(2x-3)(2x+3)}\]

2-ج) حل \(B = 55\) في \(\mathbb{N}\) :

\[4x^2-9 = 55 \Rightarrow 4x^2 = 64 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4\]

بما أن \(x \in \mathbb{N}\) : \(\boxed{x = 4}\)

2-د) حل \(4x^2+4x+1 = 2(2x+5)\) :

\[(2x+1)^2 = 2(2x+5)\]

وهذا بالضبط تعريف \(B = 0\) : \[B = (2x+1)^2-2(2x+5) = 0 \Rightarrow (2x-3)(2x+3) = 0\]

\[\boxed{x = \dfrac{3}{2} \quad \text{أو} \quad x = -\dfrac{3}{2}}\]

3-أ) حساب \(a^2\)، \(b^2\)، \(ab\) :

\[a^2 = (2+\sqrt{3})^2 = 4+4\sqrt{3}+3 = \boxed{7+4\sqrt{3}}\]

\[b^2 = (1-2\sqrt{3})^2 = 1-4\sqrt{3}+12 = \boxed{13-4\sqrt{3}}\]

\[ab = (2+\sqrt{3})(1-2\sqrt{3}) = 2-4\sqrt{3}+\sqrt{3}-2\times3 = 2-3\sqrt{3}-6 = \boxed{-4-3\sqrt{3}}\]

3-ب) حساب \(c\) و \(d\) :

حساب \(c = (3-\sqrt{3})^2\) :

\[3-\sqrt{3} = (2+\sqrt{3})+(1-2\sqrt{3}) = a+b\]

\[c = (a+b)^2 = a^2+2ab+b^2 = (7+4\sqrt{3})+2(-4-3\sqrt{3})+(13-4\sqrt{3})\] \[= 7+4\sqrt{3}-8-6\sqrt{3}+13-4\sqrt{3} = 12-6\sqrt{3}\] \[\boxed{c = 12-6\sqrt{3}}\]

تحقق : \((3-\sqrt{3})^2 = 9-6\sqrt{3}+3 = 12-6\sqrt{3}\) ✓

حساب \(d = 2\sqrt{7+4\sqrt{3}}-\sqrt{13-4\sqrt{3}}\) :

\[\sqrt{7+4\sqrt{3}} = \sqrt{a^2} = |a| = a = 2+\sqrt{3} \quad (\text{لأن } a > 0)\]

\[\sqrt{13-4\sqrt{3}} = \sqrt{b^2} = |b| = |1-2\sqrt{3}|\]

بما أنّ \(\left(2\sqrt{3}\right)^{2} = 12\) و \(1^{2} = 1\) و \(12 > 1\)، و العددان موجبان، فإنّ \(2\sqrt{3} > 1\) → \(b = 1-2\sqrt{3} < 0\) → \(|b| = 2\sqrt{3}-1\)

\[d = 2(2+\sqrt{3})-(2\sqrt{3}-1) = 4+2\sqrt{3}-2\sqrt{3}+1 = \boxed{5}\]