Question

24、已知a=1999，則|3a³-2a²+4a-1|-|3a³-3a²+3a-2001|=

4000000

．

Answer 1

分析：據(jù)絕對(duì)值的定義，由已知a=1999，可得|3a³-2a²+4a-1|=3a³-2a²+4a-1，|3a³-3a²+3a-2001|=3a³-3a²+3a-2001，去掉絕對(duì)值化簡(jiǎn)即可．

解答：解：當(dāng)a=1999時(shí)，|3a³-2a²+4a-1|=3a³-2a²+4a-1，|3a³-3a²+3a-2001|=3a³-3a²+3a-2001，
∴原式=3a³-2a²+4a-1-（3a³-3a²+3a-2001）
=a²+a+2000
=a（a+1）+2000
=1999×2000+2000
=2000×（1999+1）
=4000000．

點(diǎn)評(píng)：本題考查絕對(duì)值的化簡(jiǎn)，正數(shù)的絕對(duì)值等于它本身，0的絕對(duì)值是0，負(fù)數(shù)的絕對(duì)值為其相反數(shù)．
熟練掌握該知識(shí)點(diǎn)是解答本題的關(guān)鍵．