化簡、恆等式

(3/4) + [x + (1/2)]² = x² + x + 1 是不是恆等式？

左式 = (3/4) + [x + (1/2)]²
= (3/4) + [x² + x + (1/4)]
= (3/4) + x² + x + (1/4)
= x² + x + 1
= 右式

上式是恆等式。
（但若題目是(3/4) + [(x + 1)/2]² = x² + x + 1 則不是恆等式。）


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AB = 4cm, BC = 5cm, CD = 6cm, AD = 7cm
你想問甚麼？


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(3/4) + [x + (1/2)]² = x² + x + 1 是不是恆等式？

答案如第一題。


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(2x + 3)²
= (2x)² + 2(2x)(3) + (3)²
= 4x² + 12x + 9


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(2 - 5y)²
= (2)² - 2(2)(5y) + (5y)²
= 4 - 20y + 25y²


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(2a + 3b)²
= (2a)² + 2(2a)(3b) + (3b)²
= 4a² + 12ab + 9b²


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49 - m²
= 7² - m²
= (7 + m)(7 - m)


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100r² - 49s²
= (10r)² - (7s)²
= (10r + 7s)(10r - 7s)


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27a² - 3b² (題目 2b² 應為 3b²)
= 3(9a² - b²)
= 3[(3a)² - b²]
= 3(3a + b)(3a - b)


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-18(a + 1)²
= -18(a² + 2a + 1)
= -18a² - 32a - 32


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3x²y³/21xy⁴
= x/7y


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(2a - 4b)/(2b - a)
= -(4b - 2a)/(2b - a)
= -2(2b - a)/(2b - a)
= -2


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(3b/5a²) * (a/b)　[* = 乘號]
= 3ab/5a²b
= 3/5a


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(4x²/6y³) ÷ (12x/2y²)
= (4x²/6y³) * (2y²/12x)
= 8x²y²/72xy³
= x/9y


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(2a - 4b)/(3a + 6b) * (10b + 5a)/(6b - 3a)
= 2(a - 2b)/3(a + 2b) * 5(2b + a)/3(2b - a)
= 2(a - 2b)/3(a + 2b) * 5(a + 2b)/[-3(a -2b)
= 10(a - 2b)(a + 2b)/(-9)(a + 2b)(a - 2b)
= -10/9