⒈(x+1)/(x^2-x-6)+3/(5x^2-10+5x)-(x-2)/(x^2-4x+3)
3个回答
1、原式=(x+1)/(x-3)(x+2)+3/[5(x-1)(x+2)]-(x-2)/[(x-1)(x-3)]
=[5(x+1)(x-1)+3(x-3)-(x-2)(x+2)]/[5(x-1)(x+2)(x-3)]
=(5x^2-5+3x-9-x^2+4)/[5(x-1)(x+2)(x-3)]=(4x^2+3x-10)/[5(x-1)(x+2)(x-3)]
=(x+2)(4x-5)/[5(x-1)(x+2)(x-3)]=(4x-5)/[5(x-1)(x-3)]
2、原式=(x+a-x+a)/(x^2-a^2)-2a/(x^2+a^2)-4a^3/(x^4-a^4)
=2a(x^2+a^2-x^2+a^2)/(x^4-a^4)-4a^3/(x^4-a^4)=0
3、原式=(x+2-x+2)/(x^2-4)-4(x^2+4)-32/(x^4-16)
=4(x^2+4-X^2+4)/(x^4-16)-32/(x^4-16)=0
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