F.2 maths questions x2

1. 證明(1/4)(x - 2)(x - 6) = (x/2 - 2)² - 1 是恆等式

左式
= (1/4)(x - 2)(x - 6)
= 1/4(x² - 8x + 12)
= x²/4 - 2x + 3
= (x²/4 - 2x + 4) - 1
= [(x/2)² - 2(x/2)(2) + (2)²] - 1
= (x/2 - 2)² - 1
= 右式

故可證得 (1/4)(x - 2)(x - 6) ≡ (x/2 - 2)² - 1


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2.展開 45(9x/5 - y)(5y/9 + x)

45(9x/5 - y)(5y/9 + x)
= 5(9x/5 - y)•9(5y/9 + x)
= (9x - 5y)(5y + 9x)
= (9x - 5y)(9x + 5y)
= (9x)² - (5y)²
= 81x² - 25y²