帮忙解l两道数学题已知(6y+12)/(y^2+4y+3)=A/(y+3)+B/(y+1),求实数A,B若关于x的方程k/(x^2-x)-1/3x=1/(3x-3）有增根,求曾根和k值

Question

帮忙解l两道数学题
已知(6y+12)/(y^2+4y+3)=A/(y+3)+B/(y+1),求实数A,B
若关于x的方程k/(x^2-x) - 1/3x=1/(3x-3）有增根,求曾根和k值

Answer 1

(6y+12)/(y^2+4y+3)=A/(y+3)+B/(y+1)
(6y+12)/(y+3)(y+1)=〔A(y+1)+B(y+3)〕/(y+3)(y+1)
(6y+12)/(y+3)(y+1)=〔（A+B)y+(A+3B)〕/(y+3)(y+1)
所以 A+B=6,A+3B=12,解得A=3,B=3.
先通分,分母全部变为3X(X-1),得到(3K-X+1)/3X(X-1)=X/3X(X-1)
要得到曾根,则X=0或者1
3K-X+1=X得到3K+1=2X
若X=0,那么K=-1/3
若X=1,那么K=1/3