\[A = x^2-9 \qquad B = 2x^2-12x+18\]
1- أ- احسب \(A\) إذا \(x = -\dfrac{\sqrt{5}}{2}\) ؛ ب- فكّك \(A\)
2- بيّن أن \(B = 2(3-x)^2\)
3- فكّك \(B+A\)
4- أوجد \(x\) إذا علمت أن \(A\) و \(B\) متقابلان
1-أ) \(A = \dfrac{5}{4}-9 = \dfrac{5-36}{4} = -\dfrac{31}{4}\)
3) \(B+A = 2(3-x)^2+(x-3)(x+3) = 2(x-3)^2+(x-3)(x+3) = (x-3)\big[2(x-3)+(x+3)\big] = (x-3)(3x-3) = 3(x-3)(x-1)\)
4) \(A+B=0\) → \((x-3)(x+3)+2(x-3)^2=0\) → \((x-3)(x+3+2x-6)=0\) → \((x-3)(3x-3)=0\) → \(x=3\) أو \(x=1\)
1-أ) \[A = \left(-\dfrac{\sqrt{5}}{2}\right)^2-9 = \dfrac{5}{4}-9 = \dfrac{5-36}{4} = \boxed{-\dfrac{31}{4}}\]
1-ب) \[A = x^2-3^2 = \boxed{(x-3)(x+3)}\]
2) \[B = 2x^2-12x+18 = 2(x^2-6x+9) = 2(x-3)^2 = \boxed{2(3-x)^2} \checkmark\]
3) تفكيك \(B+A\) :\[B+A = 2(x-3)^2+(x-3)(x+3) = (x-3)\big[2(x-3)+(x+3)\big] = (x-3)(2x-6+x+3)\]
\[= (x-3)(3x-3) = 3(x-3)(x-1)\]
\[\boxed{B+A = 3(x-3)(x-1)}\]
4) \(A\) و \(B\) متقابلان :\[A+B = 0 \Rightarrow 3(x-3)(x-1) = 0\]
\[\boxed{x = 3 \quad \text{أو} \quad x = 1}\]