الجذاآت المعتبرة — التمرين 9

✅ التصحيح المفصّل

1-أ) إثبات \(A = x^2-x+\dfrac{1}{4}\) :

\[A = \left(x+\dfrac{1}{2}\right)^2-2x = x^2+2 \cdot x \cdot \dfrac{1}{2}+\dfrac{1}{4}-2x = x^2+x+\dfrac{1}{4}-2x = \boxed{x^2-x+\dfrac{1}{4}}\]

1-ب) حساب \(A\) عند \(x = \sqrt{2}-\dfrac{1}{2}\) :

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

1-ج) تفكيك \(A\) :

\[A = x^2-x+\dfrac{1}{4} = \boxed{\left(x-\dfrac{1}{2}\right)^2}\]

1-د) حل \(A = \dfrac{x^2}{2}-\dfrac{x}{2}\) :

\[\left(x-\dfrac{1}{2}\right)^2 = \dfrac{x}{2}(x-1) = \dfrac{x}{2}\left(x-1\right)\]

ننشر الطرف الأيسر : \[x^2-x+\dfrac{1}{4} = \dfrac{x^2}{2}-\dfrac{x}{2}\] \[x^2-x+\dfrac{1}{4}-\dfrac{x^2}{2}+\dfrac{x}{2} = 0\] \[\dfrac{x^2}{2}-\dfrac{x}{2}+\dfrac{1}{4} = 0\]

نضرب في 4 : \[2x^2-2x+1 = 0\]

⛔ لا داعي لأدوات خارج برنامج التاسعة : نستعمل الشكل القانوني، و هو أداة الدرس نفسها.

\[2x^{2}-2x+1 = 2\left(x-\dfrac{1}{2}\right)^{2}+\dfrac{1}{2}\]

و بما أنّ \(\left(x-\dfrac{1}{2}\right)^{2} \ge 0\) فإنّ \(2x^{2}-2x+1 \ge \dfrac{1}{2} > 0\) : المعادلة لا تقبل حلاّ في \(\mathbb{R}\) ← \(S=\emptyset\)

2-أ) تفكيك \(B-A\) :

\[B-A = 25x^2-\left(x-\dfrac{1}{2}\right)^2 = (5x)^2-\left(x-\dfrac{1}{2}\right)^2\] \[= \left[5x+\left(x-\dfrac{1}{2}\right)\right]\left[5x-\left(x-\dfrac{1}{2}\right)\right]\] \[= \left(6x-\dfrac{1}{2}\right)\left(4x+\dfrac{1}{2}\right)\] \[\boxed{= \dfrac{1}{4}(12x-1)(8x+1)}\]

2-ب) حل \(A^2-AB = AB-B^2\) :

\[A^2-AB-AB+B^2 = 0\] \[A^2-2AB+B^2 = 0\] \[(A-B)^2 = 0\] \[A = B\]

\[\left(x-\dfrac{1}{2}\right)^2 = 25x^2\] \[(5x)^2 - \left(x-\dfrac{1}{2}\right)^2 = 0\] \[\left(6x-\dfrac{1}{2}\right)\left(4x+\dfrac{1}{2}\right) = 0\]

\[x = \dfrac{1}{12} \quad \text{أو} \quad x = -\dfrac{1}{8}\]

\[\boxed{S = \left\{\dfrac{1}{12} \;;\; -\dfrac{1}{8}\right\}}\]