对所有x=-1且x≠-2的值,等式1/(x+1)(x+2)=A/(x+1) +B/(x+2)恒成立,求A、B的值
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对所有x=-1且x≠-2的值,等式1/(x+1)(x+2)=A/(x+1) +B/(x+2)恒成立,求A、B的值
应用以上结论计算:1/(1*2) +1/(2*3) +1/(3*4)...+1/(99*100)
应用以上结论计算:1/(1*2) +1/(2*3) +1/(3*4)...+1/(99*100)
A/(x+1) +B/(x+2)
=[A(x+2)+B(x+1)]/(x+1)(x+2)
=[(A+B)x+(2A+B)]/(x+1)(x+2)
=1/(x+1)(x+2)
所以(A+B)x+(2A+B)=1=0*x+1
所以A+B=0
2A+B=1
相减
A=1,B=-A=-1
1/(x+1)(x+2)
=[(x+2)-(x+1)]/(x+1)(x+2)
=(x+2)/(x+1)(x+2)-(x+1)/(x+1)(x+2)
=1/(x+1)-1/(x+2)
=1/(x+1)+(-1)/(x+2)
=A/(x+1) +B/(x+2)
所以A=1,B=-1
1/(1*2) +1/(2*3) +1/(3*4)...+1/(99*100)
=(1/1-1/2)+(1/2-1/3)+……+(1/99-1/100)
=1-1/100
=99/100
=[A(x+2)+B(x+1)]/(x+1)(x+2)
=[(A+B)x+(2A+B)]/(x+1)(x+2)
=1/(x+1)(x+2)
所以(A+B)x+(2A+B)=1=0*x+1
所以A+B=0
2A+B=1
相减
A=1,B=-A=-1
1/(x+1)(x+2)
=[(x+2)-(x+1)]/(x+1)(x+2)
=(x+2)/(x+1)(x+2)-(x+1)/(x+1)(x+2)
=1/(x+1)-1/(x+2)
=1/(x+1)+(-1)/(x+2)
=A/(x+1) +B/(x+2)
所以A=1,B=-1
1/(1*2) +1/(2*3) +1/(3*4)...+1/(99*100)
=(1/1-1/2)+(1/2-1/3)+……+(1/99-1/100)
=1-1/100
=99/100
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