°1 \(A = (x-2)(x-1)\)
أ- أنشر واختصر \(A\) ؛ ب- أحسب \(A\) إذا \(x=1\) ثم \(x=1+\sqrt{2}\)
°2 \(B = (1-x)^2(x+1)+(x-1)\)
أ- أحسب \(B\) إذا \(x = 1+\sqrt{2}\) ؛ ب- بيّن أن \(B = x^2(x-1)\)
°3 \(C = A+B\)
أ- بيّن أن \((x-1)(x+2) = x^2+x-2\)
ب- استنتج أن \(C = (x-1)^2(x+2)\)
ج- أحسب \(C\) بطريقتين إذا \(x = 1+\sqrt{2}\)
°4 أوجد \(x\) إذا \(C = x+2\)
°1-أ) \(A = x^2-3x+2\)
°1-ب) \(x=1\) : \(A=0\) ؛ \(x=1+\sqrt{2}\) : \(A = (1+\sqrt{2}-2)(1+\sqrt{2}-1) = (-1+\sqrt{2})(\sqrt{2}) = 2-\sqrt{2}\)
°2-أ) \(B = (1-1-\sqrt{2})^2(2+\sqrt{2})+(1+\sqrt{2}-1) = 2(2+\sqrt{2})+\sqrt{2} = 4+2\sqrt{2}+\sqrt{2} = 4+3\sqrt{2}\)
°2-ب) \(B = (x-1)^2(x+1)+(x-1) = (x-1)\big[(x-1)(x+1)+1\big] = (x-1)(x^2-1+1) = x^2(x-1)\)
°3-ب) \(C = (x-2)(x-1)+x^2(x-1) = (x-1)(x-2+x^2) = (x-1)(x^2+x-2) = (x-1)(x-1)(x+2) = (x-1)^2(x+2)\)
°4) \((x-1)^2(x+2)-(x+2)=0\) → \((x+2)\left[(x-1)^2-1\right]=0\) → \((x+2)(x-2)x=0\)
°1-أ) \[A = (x-2)(x-1) = x^2-x-2x+2 = \boxed{x^2-3x+2}\]
°1-ب) \(x=1\) : \(A = 1-3+2 = \boxed{0}\)
\(x=1+\sqrt{2}\) : \(A = (\sqrt{2}-1)(\sqrt{2}) = 2-\sqrt{2}\) ← \(\boxed{A = 2-\sqrt{2}}\)
°2-أ) \(x-1 = \sqrt{2}\), \((1-x)^2 = \sqrt{2}^2 = 2\), \(x+1 = 2+\sqrt{2}\)
\[B = 2(2+\sqrt{2})+\sqrt{2} = 4+2\sqrt{2}+\sqrt{2} = \boxed{4+3\sqrt{2}}\]
°2-ب) \[(1-x)^2(x+1)+(x-1) = (x-1)^2(x+1)+(x-1) = (x-1)\big[(x-1)(x+1)+1\big]\]
\[= (x-1)(x^2-1+1) = x^2(x-1) \checkmark\]
°3-أ) \[(x-1)(x+2) = x^2+2x-x-2 = x^2+x-2 \checkmark\]
°3-ب) \[C = A+B = (x^2-3x+2)+x^2(x-1) = x^3-x^2+x^2-3x+2-... \]
الأسهل :
\[C = (x-2)(x-1)+x^2(x-1) = (x-1)(x-2+x^2) = (x-1)(x^2+x-2)\]
من °3-أ) : \(x^2+x-2 = (x-1)(x+2)\)
\[\boxed{C = (x-1)^2(x+2)}\]
°3-ج) بطريقتين (\(x=1+\sqrt{2}\)) :الطريقة 1 : \(C = A+B = (2-\sqrt{2})+(4+3\sqrt{2}) = \boxed{6+2\sqrt{2}}\)
الطريقة 2 : \(C = (x-1)^2(x+2) = (\sqrt{2})^2(3+\sqrt{2}) = 2(3+\sqrt{2}) = 6+2\sqrt{2}\) ✓
°4) حل \(C = x+2\) :\[(x-1)^2(x+2) = x+2\]
\[(x+2)\left[(x-1)^2-1\right] = 0\]
\[(x+2)(x-1+1)(x-1-1) = 0\]
\[(x+2)(x)(x-2) = 0\]
\[\boxed{x = -2 \quad \text{أو} \quad x = 0 \quad \text{أو} \quad x = 2}\]