الجذاآت المعتبرة — التمرين 7

✅ التصحيح المفصّل

1) نشر واختصار \(A\) :

\[(x\sqrt{3}+1)^2 = (\sqrt{3})^2 x^2 + 2\sqrt{3}x + 1 = 3x^2 + 2\sqrt{3}x + 1\]

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

\[A = (3x^2+2\sqrt{3}x+1) - (x^2-2\sqrt{3}x+3)\] \[\boxed{A = 2x^2 + 4\sqrt{3}x - 2}\]

نشر واختصار \(B\) :

\[B = (x\sqrt{2}+1)(x\sqrt{2}-1) = (x\sqrt{2})^2 - 1^2 = 2x^2 - 1\] \[\boxed{B = 2x^2 - 1}\]

2) تفكيك \(A\) إلى جذاء عوامل :

\[A = 2x^2 + 4\sqrt{3}x - 2 = 2(x^2 + 2\sqrt{3}x - 1)\]

نُكمل المربع : \[x^2 + 2\sqrt{3}x - 1 = (x+\sqrt{3})^2 - (\sqrt{3})^2 - 1 = (x+\sqrt{3})^2 - 3 - 1 = (x+\sqrt{3})^2 - 4\]

\[(x+\sqrt{3})^2 - 4 = (x+\sqrt{3}+2)(x+\sqrt{3}-2)\]

إذًا : \[\boxed{A = 2(x+\sqrt{3}+2)(x+\sqrt{3}-2)}\]

3) مجموعة حلول \(A = 0\) :

\[2(x+\sqrt{3}+2)(x+\sqrt{3}-2) = 0\]

\[x + \sqrt{3} + 2 = 0 \quad \text{أو} \quad x + \sqrt{3} - 2 = 0\]

\[\boxed{S = \{-\sqrt{3}-2 \;;\; 2-\sqrt{3}\}}\]

4) مجموعة حلول \(A\) مقابل \(B\) :

مقابل يعني : \(A + B = 0\)

\[(2x^2 + 4\sqrt{3}x - 2) + (2x^2 - 1) = 0\] \[4x^2 + 4\sqrt{3}x - 3 = 0\]

نُكمل المربع : \[4\left(x^2 + \sqrt{3}x\right) = 3\] \[4\left[\left(x+\dfrac{\sqrt{3}}{2}\right)^2 - \dfrac{3}{4}\right] = 3\] \[4\left(x+\dfrac{\sqrt{3}}{2}\right)^2 - 3 = 3\] \[4\left(x+\dfrac{\sqrt{3}}{2}\right)^2 = 6\] \[\left(x+\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{6}{4} = \dfrac{3}{2}\]

\[x+\dfrac{\sqrt{3}}{2} = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2} \quad \text{أو} \quad x+\dfrac{\sqrt{3}}{2} = -\dfrac{\sqrt{6}}{2}\]

\[x = \dfrac{\sqrt{6}-\sqrt{3}}{2} \quad \text{أو} \quad x = \dfrac{-\sqrt{6}-\sqrt{3}}{2}\]

\[\boxed{S = \left\{\dfrac{\sqrt{6}-\sqrt{3}}{2} \;;\; \dfrac{-\sqrt{6}-\sqrt{3}}{2}\right\}}\]