计算：a-3分之1减a的平方-9分之6减6+2a分之a-1
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计算：a-3分之1减a的平方-9分之6减6+2a分之a-1
计算：a-3分之1减a的平方-9分之6减6+2a分之a-1
计算：a-3分之1减a的平方-9分之6减6+2a分之a-1
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1/(a-3)-6/(a^2-9)-(a-1)/(6+2a)
=1/(a-3)-6/[(a-3)(a+3)]-(a-1)/[2(a+3)]
=2(a+3)/[2(a-3)(a+3)]-6*2/[2(a-3)(a+3)]-(a-1)(a-3)/[2(a-3)(a+3)]
=[2(a+3)-12-(a-1)(a-3)]/[2(a-3)(a+3)]
=[2a+6-12-(...
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1/(a-3)-6/(a^2-9)-(a-1)/(6+2a)
=1/(a-3)-6/[(a-3)(a+3)]-(a-1)/[2(a+3)]
=2(a+3)/[2(a-3)(a+3)]-6*2/[2(a-3)(a+3)]-(a-1)(a-3)/[2(a-3)(a+3)]
=[2(a+3)-12-(a-1)(a-3)]/[2(a-3)(a+3)]
=[2a+6-12-(a^2-4a+3)]/[2(a-3)(a+3)]
=[2a-6-a^2+4a-3]/[2(a-3)(a+3)]
=[-a^2+2a+4a-6-3]/[2(a-3)(a+3)]
=[-a^2+6a-9]/[2(a-3)(a+3)]
=-(a^2-6a+9)/[2(a-3)(a+3)]
=-(a-3)^2/[2(a-3)(a+3)]
=-(a-3)/[2(a+3)]
=-(a-3)/(2a+6)
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你这样问很有歧义呀，至少加括号嘛，否则我们没法帮你算的~~
1/(a-3)-6/(a-3)(a+3)-(a-1)/2(a+3)
=[2(a+3)-12-(a-1)(a-3)]/[2(a-3)(a+3)]
=[2(a-3)-(a-1)(a-3)]/[2(a-3)(a+3)]
=(a-3)(2-a+1)/[2(a-3)(a+3)]
=(3-a)/3(a+3)
6/(a^2-9)=1/(a-3)-1/(a+3)