التمرين رقم 2

✅ الحل الكامل

\[(x-\sqrt{2})^2 = x^2-2\sqrt{2}x+2\]

\[(2x-1)^2 = 4x^2-4x+1\]

\[(\sqrt{3}x-2\sqrt{2})^2 = 3x^2-4\sqrt{6}x+8\]

\[(\sqrt{2}x+2)^2 = 2x^2+4\sqrt{2}x+4\]

\[(x+3\sqrt{2})^2 = x^2+6\sqrt{2}x+18\]

\[(x-4)(x+4) = x^2-16\]

\[(\sqrt{2}x-1)(\sqrt{2}x+1) = 2x^2-1\]

\[(2x-3\sqrt{3})\left(2x+\dfrac{\sqrt{3}}{3}\right) = 4x^2+\dfrac{2\sqrt{3}}{3}x-6\sqrt{3}x-3\]

\[= 4x^2+x\left(\dfrac{2\sqrt{3}}{3}-6\sqrt{3}\right)-3 = 4x^2+x\cdot\dfrac{2\sqrt{3}-18\sqrt{3}}{3}-3 = 4x^2-\dfrac{16\sqrt{3}}{3}x-3\]

\[x^2-1 = (x-1)(x+1)\]

\[x^2-9 = (x-3)(x+3)\]

\[x^2-2 = (x-\sqrt{2})(x+\sqrt{2})\]

\[2x^2-1 = (\sqrt{2}x-1)(\sqrt{2}x+1)\]

\[3x^2-2 = (\sqrt{3}x-\sqrt{2})(\sqrt{3}x+\sqrt{2})\]

\[64x^2-36 = 4(16x^2-9) = 4(4x-3)(4x+3)\]

\[x^2+2x+1 = (x+1)^2\]

\[x^2-2x+1 = (x-1)^2\]

\[x^2-4x+4 = (x-2)^2\]

\[4x^2-3 = (2x-\sqrt{3})(2x+\sqrt{3})\]

\[7x^2-10 = (\sqrt{7}x-\sqrt{10})(\sqrt{7}x+\sqrt{10})\]

\[25x^2+20x+4 = (5x+2)^2\]

\[4t^2+2t+\dfrac{1}{4} = \left(2t+\dfrac{1}{2}\right)^2\]

\[2y^2+2\sqrt{6}y+3 = (\sqrt{2}y+\sqrt{3})^2\]

\[(3x+4)(2x-3)-4x^2+9 = (3x+4)(2x-3)-(2x-3)(2x+3) = (2x-3)(3x+4-2x-3) = (2x-3)(x+1)\]

\[A = 4x(5+x^2)(1-x^2)+1+x^4-2\]

\[= 4x(5+x^2)(1-x^2)+(x^4-1)\]

\[= 4x(5+x^2)(1-x^2)+(x^2-1)(x^2+1)\]

\[= (x^2-1)\big[-4x(5+x^2)+(x^2+1)\big]\]

انتبه : \(1-x^2 = -(x^2-1)\)

\[= (x^2-1)\big[x^2+1-4x(5+x^2)\big]\]

\[= (x-1)(x+1)(x^2+1-20x-4x^3)\]

هذا التفكيك معقد — الأرجح أن \(A = 4x(5+x^2)(1-x^2)+(1+x^4-2)\) يُقصد به \((x^2-1)^2\) من باب التحليل المعرفي. نكتفي بـ :

\[A = (x^2-1)\big[(x^2+1)-4x(5+x^2)\big] = (x-1)(x+1)(-4x^3+x^2-20x+1)\]

\[B = x^2-9-(x+3)(2x-5) = (x-3)(x+3)-(x+3)(2x-5) = (x+3)(x-3-2x+5) = \boxed{(x+3)(2-x)}\]

\[C = (3x+1)^2-4 = (3x+1-2)(3x+1+2) = \boxed{(3x-1)(3x+3)} = 3(3x-1)(x+1)\]

\[D = 16-(2x-1)^2 = (4-2x+1)(4+2x-1) = \boxed{(5-2x)(3+2x)}\]