又來問國二數學了~這次有3題要麻煩大家 謝謝^^?

1(1)
f(x)=(x+2)(4x^2-2x-6)+4
f(x)=4x^3-2x^2-6x+8x^2-4x-12+4
f(x)=4x^3+6x^2-10x-8
根據長除法，你會計算到q(x)=2x^2+2x-6

1(2)根據長除法，你會計算到a=-2

2
上底=HG=3
下底=AD=2x
高=AG=2x
五邊形AGHCD面積
=(3+2x)*2x/2
=3x+2x^2
=2x^2+3x

3(1)
[(xy + 2)(xy ‒ 2) ‒ 2(x²y² ‒ 2)] ÷ xy
= (x²y² ‒ 4 ‒ 2x²y² + 4) ÷ xy
= (-x²y²) ÷ xy
= -xy

3(2)
代x=10及y=-1/25入(1)
-10*-1/25
=2/5

1.
f(x) = (x + 2)(4x² ‒ 2x ‒ 6) + 4 ...... [1]
f(x) = (2x + 1)q(x) + a ...... [2]

[2] = [1]
(2x + 1)q(x) + a = (x + 2)(4x² ‒ 2x ‒ 6) + 4
(2x + 1)q(x) + a = 4x³ + 6x² ‒ 10x ‒ 8
(2x + 1)q(x) + a = (4x³ + 2x²) + (4x² + 2x) + (‒12x ‒ 6) ‒ 2
(2x + 1)q(x) + a = 2x²(2x + 1) + 2x(2x + 1) ‒6(2x + 1) ‒ 2
(2x + 1)q(x) + a = (2x + 1)(2x² + 2x ‒6) ‒ 2

Ans:
(1) q(x) = 2x² + 2x ‒ 6
(2) a = ‒2


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2.
五邊形AGHCD面積
= 梯形EHCD面積 + 長方形AGHE面積
= (1/2) × (2x + 3) × (2x ‒ 3) + (2x) × 3
= (1/2)(4x² ‒ 9) + 6x
= 2x² + 6x ‒ (9/2) 平方單位


====
3.
(1)
[(xy + 2)(xy ‒ 2) ‒ 2(x²y² ‒ 2)] ÷ xy
= [x²y² ‒ 4 ‒ 2x²y² + 4] ÷ xy
= [‒ 2x²y²] ÷ xy
= ‒2xy

(2)
(1) 式之值
= ‒10 × (‒1/25)
= 2/5