\[A = -8x^2+6x-1\]
1- أحسب \(A\) في حالة \(x = -\dfrac{\sqrt{2}}{4}\)
2- أ- بيّن أن \(x^2-A = (3x-1)^2\) ؛ ب- استنتج أن \(A = (-2x+1)(4x-1)\)
3- لتكن \(B = A-4x^2+2x\)
أ- أثبت أن \(B = (-2x+1)(6x-1)\) ؛ ب- أوجد \(x\) بحيث \(A = 4x^2-2x\)
1) \(A = -8\left(\dfrac{\sqrt{2}}{4}\right)^2+6\left(-\dfrac{\sqrt{2}}{4}\right)-1 = -8 \cdot \dfrac{2}{16}-\dfrac{6\sqrt{2}}{4}-1 = -1-\dfrac{3\sqrt{2}}{2}-1 = -2-\dfrac{3\sqrt{2}}{2}\)
2-أ) \(x^2-A = x^2+8x^2-6x+1 = 9x^2-6x+1 = (3x-1)^2\) ✓
2-ب) \(A = x^2-(3x-1)^2 = (x-3x+1)(x+3x-1) = (-2x+1)(4x-1)\)
3-أ) \(B = (-2x+1)(4x-1)-4x^2+2x\)
\(= (-2x+1)(4x-1)+2x(1-2x)\)
\(= (-2x+1)(4x-1)+2x(-2x+1)\)
\(= (-2x+1)(4x-1+2x)\)
\(= (-2x+1)(6x-1)\)
3-ب) \(A = 4x^2-2x\) → \(B = 0\) → \((-2x+1)(6x-1) = 0\)
\[A = -8\left(\dfrac{\sqrt{2}}{4}\right)^2+6\left(-\dfrac{\sqrt{2}}{4}\right)-1 = -8 \cdot \dfrac{2}{16}-\dfrac{6\sqrt{2}}{4}-1 = -1-\dfrac{3\sqrt{2}}{2}-1 = \boxed{-2-\dfrac{3\sqrt{2}}{2}}\]
2-أ) إثبات \(x^2-A = (3x-1)^2\) :\[x^2-A = x^2-(-8x^2+6x-1) = x^2+8x^2-6x+1 = 9x^2-6x+1 = (3x)^2-2\cdot3x\cdot1+1^2 = (3x-1)^2 \checkmark\]
2-ب) استنتاج \(A = (-2x+1)(4x-1)\) :\[A = x^2-(3x-1)^2 = \big[x-(3x-1)\big]\big[x+(3x-1)\big] = (-2x+1)(4x-1) \checkmark\]
3-أ) إثبات \(B = (-2x+1)(6x-1)\) :\[B = A-4x^2+2x = (-2x+1)(4x-1)-4x^2+2x\]
\[= (-2x+1)(4x-1)+2x(1-2x)\]
\[= (-2x+1)(4x-1)+2x(-2x+1)\]
\[= (-2x+1)\big[(4x-1)+2x\big]\]
\[= \boxed{(-2x+1)(6x-1)}\]
3-ب) حل \(A = 4x^2-2x\) :\[A-4x^2+2x = 0 \Rightarrow B = 0\]
\[(-2x+1)(6x-1) = 0\]
\[\boxed{x = \dfrac{1}{2} \quad \text{أو} \quad x = \dfrac{1}{6}}\]