试题

题目:
若m=x3-3x2y+2xy2+3y3,n=x3-2x2y+xy2-5y3,则2x3-7x2y+5xy2+14y3的值为(  )



答案
C
解:∵m=x3-3x2y+2xy2+3y3,n=x3-2x2y+xy2-5y3
∴3m-n=3(x3-3x2y+2xy2+3y3)-(x3-2x2y+xy2-5y3
=3x3-9x2y+6xy2+9y3-x3+2x2y-xy2+5y3
=2x3-7x2y+5xy2+14y3
即2x3-7x2y+5xy2+14y3的值为3m-n.
故选C.
考点梳理
整式的加减.
根据各选项的式子,分别把m=x3-3x2y+2xy2+3y3,n=x3-2x2y+xy2-5y3代入,再去括号合并同类项,看结果等于2x3-7x2y+5xy2+14y3即可.
本题主要考查整式的加减,解题的关键是正确去括号,合并同类项.
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