解两个题 1.(x+2\ x^2-2x - x-1\x^2-4x+4)除以x-4\x 2.x^2-2x\x^2-1 除以
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解两个题 1.(x+2\ x^2-2x - x-1\x^2-4x+4)除以x-4\x 2.x^2-2x\x^2-1 除以(x-1 - 2x-1\x+1)
1.[(x+2)/(x^2-2x) - (x-1)/(x^2-4x+4)]/[(x-4)/x ]
=.[(x+2)/x(x-2)- (x-1)/(x-2)^2]*x /(x-4)
=[(x+2)(x-2)/x(x-2)^2- x(x-1)/x(x-2)^2]*x /(x-4)
=[(x^-4-x^+x)/x(x-2)^2]*x /(x-4)
=(x-4)/x(x-2)^2*x /(x-4)
=1/(x-2)^2
2.[(x^2-2x)/(x^2-1)] /[x-1 - (2x-1)/(x+1)]
=[(x^2-2x)/(x-1)(x+1)]/[(x^2-1- 2x+1)/(x+1)]
=[(x^2-2x)/(x-1)(x+1)]/[(x^2- 2x)/(x+1)]
==[(x^2-2x)/(x-1)(x+1)]*[(x+1)/(x^2- 2x)]
=1/(x-1)
=.[(x+2)/x(x-2)- (x-1)/(x-2)^2]*x /(x-4)
=[(x+2)(x-2)/x(x-2)^2- x(x-1)/x(x-2)^2]*x /(x-4)
=[(x^-4-x^+x)/x(x-2)^2]*x /(x-4)
=(x-4)/x(x-2)^2*x /(x-4)
=1/(x-2)^2
2.[(x^2-2x)/(x^2-1)] /[x-1 - (2x-1)/(x+1)]
=[(x^2-2x)/(x-1)(x+1)]/[(x^2-1- 2x+1)/(x+1)]
=[(x^2-2x)/(x-1)(x+1)]/[(x^2- 2x)/(x+1)]
==[(x^2-2x)/(x-1)(x+1)]*[(x+1)/(x^2- 2x)]
=1/(x-1)
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