Question

VA ROG AJUTATI MA ESTE FOARTE URGENT,DAU COROANA!!!!

Answer 1

$(3\sqrt{2}x + 2)^{2} = (3\sqrt{2}x)^{2} + 2 \cdot 3\sqrt{2}x \cdot 2 + 2^{2} = 18x^{2} + 12\sqrt{2}x + 4$

$(2\sqrt{2}x - 5)^{2} = (2\sqrt{2}x)^{2} - 2 \cdot 2\sqrt{2}x \cdot 5 + 5^{2} = 8x^{2} - 20\sqrt{2} + 25$

$(3x - 2\sqrt{2})^{2} = (3x)^{2} - 2 \cdot 3x \cdot 2\sqrt{2} + (2\sqrt{2})^{2} = 9x^{2} - 12 \sqrt{2}x + 8$

$(5\sqrt{2}x + 1)(\sqrt{2}x - 1) = 5\sqrt{2}x \cdot \sqrt{2}x - 5\sqrt{2}x + \sqrt{2}x - 1 = 10x^{2} - 4\sqrt{2}x - 1$

⇒

$(3\sqrt{2}x + 2)^{2} - (2\sqrt{2}x - 5)^{2} + (3x - 2\sqrt{2})^{2} - (5\sqrt{2}x + 1)(\sqrt{2}x - 1) =$

$= 18x^{2} + 12\sqrt{2}x + 4 - (8x^{2} - 20\sqrt{2} + 25) + (9x^{2} - 12 \sqrt{2}x + 8) - (10x^{2} - 4\sqrt{2}x - 1)$

$= 18x^{2} + 12\sqrt{2}x + 4 - 8x^{2} + 20\sqrt{2} - 25 + 9x^{2} - 12 \sqrt{2}x + 8 - 10x^{2} + 4\sqrt{2}x + 1$

$= (18x^{2} - 8x^{2} + 9x^{2} - 10x^{2}) + (12\sqrt{2}x + 20\sqrt{2} - 12 \sqrt{2}x + 4\sqrt{2}x) + (4 - 25 + 8 + 1)$

$= 9x^{2} + 24\sqrt{2} - 12$