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f(x) = (-5+3x)/[(1-x)(3-x)] = -1/(1-x) - 2/(3-x)
(1) f(x) = -1/(1-x) - (2/3)/(1-x/3)
= -∑<n=0, ∞>x^n - (2/3)∑<n=0, ∞>(x/3)^n
= -∑<n=0, ∞>[1+2/3^(n+1)]x^n
收敛域 -1 < x < 1, -1 < x/3 < 1, 解得 -1 < x < 1。
(2) f(x) = -1/[3-(x+2)] - 2/[5-(x+2)]
= -(1/3)/[1-(x+2)/3] - (2/5)/[1-(x+2)/5]
= -(1/3)∑<n=0, ∞>[(x+2)/3]^n - (2/5)∑<n=0, ∞>[(x+2)/5]^n
= -∑<n=0, ∞>[1/3^(n+1)+2/5^(n+1)](x+2)^n
收敛域 -1 < (x+2)/3 < 1, -1 < (x+2)/5 < 1, 解得 -5 < x < 1。
(1) f(x) = -1/(1-x) - (2/3)/(1-x/3)
= -∑<n=0, ∞>x^n - (2/3)∑<n=0, ∞>(x/3)^n
= -∑<n=0, ∞>[1+2/3^(n+1)]x^n
收敛域 -1 < x < 1, -1 < x/3 < 1, 解得 -1 < x < 1。
(2) f(x) = -1/[3-(x+2)] - 2/[5-(x+2)]
= -(1/3)/[1-(x+2)/3] - (2/5)/[1-(x+2)/5]
= -(1/3)∑<n=0, ∞>[(x+2)/3]^n - (2/5)∑<n=0, ∞>[(x+2)/5]^n
= -∑<n=0, ∞>[1/3^(n+1)+2/5^(n+1)](x+2)^n
收敛域 -1 < (x+2)/3 < 1, -1 < (x+2)/5 < 1, 解得 -5 < x < 1。
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