第1个回答 2019-09-23
A(1+t^2)+(Bt+C)(1+2t)≡1
t=-1/2
(5/4)A =1
A=4/5
coef. of t^2
A+2B =0
4/5 +2B=0
B= -2/5
coef. of constant
A+C =1
4/5+C=1
C=1/5
(A,B,C)=(4/5, -2/5, 1/5)本回答被网友采纳
t=-1/2
(5/4)A =1
A=4/5
coef. of t^2
A+2B =0
4/5 +2B=0
B= -2/5
coef. of constant
A+C =1
4/5+C=1
C=1/5
(A,B,C)=(4/5, -2/5, 1/5)本回答被网友采纳
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