التمرين السادس

✅ الحل الكامل والمفصل

1. تفكيك العبارات إلى جذاء عوامل

• $A = (x-3)(-x+6) - (3-x)x = (x-3)(-x+6) + (x-3)x = (x-3)[(-x+6)+x] = (x-3)(6) = 6(x-3)$.
• $B = (x-5)(x+2) + 2(x+2) = (x+2)[(x-5)+2] = (x+2)(x-3)$.
• $C = (3x+1)(x-\sqrt{2}) - (x-\sqrt{2}) = (x-\sqrt{2})[(3x+1)-1] = (x-\sqrt{2})(3x) = 3x(x-\sqrt{2})$.
• $D = 2 - \sqrt{2}x = \sqrt{2}(\sqrt{2}-x)$.

2. جد العدد الحقيقي x في كل حالة

• أ) $A=B$:
  $6(x-3) = (x+2)(x-3) \implies 6(x-3) - (x+2)(x-3) = 0$
  $(x-3)[6-(x+2)] = 0 \implies (x-3)(4-x) = 0$
  إذن $x=3$ أو $x=4$.
• ب) A و B متقابلان ($A+B=0$):
  $6(x-3) + (x+2)(x-3) = 0 \implies (x-3)[6+(x+2)] = 0$
  $(x-3)(x+8) = 0$
  إذن $x=3$ أو $x=-8$.
• ج) $|D|=\sqrt{2}$:
  $|2-\sqrt{2}x|=\sqrt{2}$.
  إما $2-\sqrt{2}x = \sqrt{2} \implies 2-\sqrt{2} = \sqrt{2}x \implies x = \frac{2-\sqrt{2}}{\sqrt{2}} = \sqrt{2}-1$.
  أو $2-\sqrt{2}x = -\sqrt{2} \implies 2+\sqrt{2} = \sqrt{2}x \implies x = \frac{2+\sqrt{2}}{\sqrt{2}} = \sqrt{2}+1$.
• د) $C=\sqrt{18}D$:
  $3x(x-\sqrt{2}) = \sqrt{18}(2-\sqrt{2}x) = 3\sqrt{2}\sqrt{2}(\sqrt{2}-x) = 6(\sqrt{2}-x)$.
  $3x(x-\sqrt{2}) = -6(x-\sqrt{2}) \implies 3x(x-\sqrt{2}) + 6(x-\sqrt{2}) = 0$.
  $(x-\sqrt{2})(3x+6) = 0$.
  إذن $x=\sqrt{2}$ أو $x=-2$.