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(1) (1/x-1) - (1/x+1) - (2/x^2+1) - (4/x^4+1)

= [(1/x-1) - (1/x+1)] - (2/x^2+1) - (4/x^4+1)

= 2/(x^2-1) - (2/x^2+1) - (4/x^4+1)

= [2/(x^2-1) - (2/x^2+1)] - (4/x^4+1)

= 4/(x^4-1) -4/(x^4+1)

=8/(x^8-1)

平方差公式要活用

(2) (3x-2/x^2-5x+6) - (2x-1/x^2-3x+2)

= (3x-2)/[(x-2)(x-3)] - (2x-1)/[(x-2)(x-1)]

= [(3x-2)(x-1) - (2x-1)(x-3)]/[(x-1)(x-2)(x-3)]

= (x^2+2x-1)/[(x-1)(x-2)(x-3)]

(3) (x^2+2x+1/x-3) * (x^2+5x-6/x^2-1) ÷ (x^3+1/x^2-5x+6)

= (x+1)^2/(x-3) * [(x-1)(x+6)]/[(x+1)(x-1)] * [(x-2)(x-3)]/[(x+1)(x^2-x+1)]

= [(x+1)^2(x-1)(x+6)(x-2)(x-3)]/[(x-3)(x+1)(x-1)(x+1)(x^2-x+1)]

= [(x+6)(x-2)]/[(x^2-x+1)]

= (x^2+4x-12)/(x^2-x+1)

(4) [x/(1-x)+(1+x)/x] / [x/(1+x)-(1+x)/x]→分數-分數分之分數+分數

= {[x^2+(1+x)(1-x)]/[x(1-x)]} / {[x^2+1+2x+x^2]/[x(1+x)]}

= {[x^2+(1+x)(1-x)]/[x(1-x)]} * {[x(1+x)]/[x^2+1+2x+x^2]}

= {1/[x(1-x)]}*[x(1+x)]/[2x^2+2x+1]

=[x(1+x)]/[x(1-x)(2x^2+2x+1)]

=(1+x)/[(1-x)(2x^2+2x+1)]

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