已知(a-c)^2-4(a-b)(b-c)=0,求证2b=a+c
已知(a-c)??-4(a-b)(b-c)=0,求证2b=a+c解:因为(a-c)??-4(a-b)(b-c)=0故:a??-2ac+c??-4ab+4ac+4b??-4bc=0即:a??+c??+4b??+2ac-4ab-4bc=0即:(a+c-2b)??=0故:a+c-2b=0故:2b=a+c 2.如果a??+b??=1,c??+d??=1,且ac+bd=0,试证明ab+cd=0解:因为a??+b??=1,c??+d??=1故:a??=1-b??,d??=1-c??因为ac+bd=0故:ac=-bd故:a??c??=b??d??故:(1-b??)c??=b??(1-c??)故:c??=b??同理:a??=d??故:a??b??=c??d??因为ac+bd=0,对于ac、bd中一正一负,即:a、b、c、d中有奇数个负数(一正三负、三正一负)或全等于0故:ab=-cd故:ab+cd=0 3.计算(a-1) ?? (a??+a+1) ?? (a+1) ?? (a??-a+1) ??=[(a-1) (a??+a+1) ]??[ (a+1) (a??-a+1)] ??=(a??-1)?? (a??+1) ??=[(a??-1)(a??+1) ]??=(a^6-1)??=a^12-2a^6+1 4.已知a+b+c=4,ab+bc+ac=4,求a??+b??+c??的值解:因为a+b+c=4故:(a+b+c) ??=4??故:a??+b??+c??+2ab+2bc+2ac=16因为ab+bc+ac=4故:a??+b??+c??=8
(??为^2因为我的电脑不好打平方)
2024-10-13 广告