Question

解方程1\x^+3x+2 + 1\x^2+5x+6 + 1\x^2+7x+12 + 1\x^2+x=4/21的解是

Answer 1

1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+x)=4/21

1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/x(x+1)=4/21

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=4/21

1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=4/21

1/x-1/(x+4)=4/21

(x+4-x)/x(x+4)=4/21

4/x(x+4)=4/21

1/x(x+4)=1/21

x(x+4)=21

x^2+4x-21=0

(x+7)(x-3)=0

x=-7或x=3

经检验x=-7或x=3是方程的解