m/1-m^2=1-n^2/-n^3+n=约分3x/6x^2za+b/a^2-b^2x^2+2x+1/x^2+x3x-6

飞天肥田2022-10-04 11:39:541条回答

m/1-m^2=
1-n^2/-n^3+n=
约分
3x/6x^2z
a+b/a^2-b^2
x^2+2x+1/x^2+x
3x-6/x^2-4x+4
12a^2b^3÷(-8a^3b)
(m^2-2m+1)÷(m^2-1)

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shinhwa0509 共回答了15个问题 | 采纳率93.3%: m/1-m^2=1+(1/m+1)
1-n^2/-n^3+n=-1/n
3x/6x^2z=1/2xz
a+b/a^2-b^2=1/a-b
x^2+2x+1/x^2+x=1+1/x
3x-6/x^2-4x+4=3/x-2
12a^2b^3÷(-8a^3b)=-3b^2/2a
(m^2-2m+1)÷(m^2-1)=m-1/m+1; 1年前