ab=cd=k⇒a=bk;c=dkab=cd=k⇒a=bk;c=dk
Suy ra: 2a+13b3a−7b=2bk+13b3bk−7b=b.(2k+13)b.(3k−7)=2k+133k−72a+13b3a−7b=2bk+13b3bk−7b=b.(2k+13)b.(3k−7)=2k+133k−7
2c+13d3c−7d=2dk+13d3dk−7d=d.(2k+13)d.(3k−7)=2k+133k−72c+13d3c−7d=2dk+13d3dk−7d=d.(2k+13)d.(3k−7)=2k+133k−7
Vậy 2a+13b3a−7b=2c+13d3c−7d2a+13b3a−7b=2c+13d3c−7d khi: ab=cd