已知abcd=1,求ax/(abc+ab+a+1)+bx/(bcd+bc+b+1)+cx/(cda+cd+c+1)+dx/(dab_数学解答

已知abcd=1,求ax/(abc+ab+a+1)+bx/(bcd+bc+b+1)+cx/(cda+cd+c+1)+dx/(dab+da+d+1)=1,求x得解

人气：448 ℃　时间：2019-10-19 18:41:49

解答

dx/(dab+da+d+1)=cdx/(cdab+cda+cd+c)=cdx/(cda+cd+c+1)=bcdx/(bcd+bc+b+1)=abcdx/(abc+ab+a+1)=x/(abc+ab+a+1),类似地,cx/(cda+cd+c+1)=abcx/(abc+ab+a+1),bx/(bcd+bc+b+1)=abx/(abc+ab+a+1),原方程成为ax/(abc+ab+...