试题

题目:
先化简再求值:3(y+1)2-5(y+1)(y-1)+2(y-1)2,其中y=-
1
2

答案
解:3(y+1)2-5(y+1)(y-1)+2(y-1)2
=3(y+1)2-3(y+1)(y-1)+2(y-1)2-2(y+1)(y-1)
=3(y+1)(y+1-y+1)+2(y-1)(y-1-y-1)
=6(y+1)-4(y-1)
=6y+6-4y+4
=2y+10,
当y=-
1
2
时,
原式=2×(-
1
2
)+10=9.
解:3(y+1)2-5(y+1)(y-1)+2(y-1)2
=3(y+1)2-3(y+1)(y-1)+2(y-1)2-2(y+1)(y-1)
=3(y+1)(y+1-y+1)+2(y-1)(y-1-y-1)
=6(y+1)-4(y-1)
=6y+6-4y+4
=2y+10,
当y=-
1
2
时,
原式=2×(-
1
2
)+10=9.
考点梳理
整式的混合运算—化简求值.
用提取公因式法将代数式降元,然后合并同类项,再代入y的值计算.
本题主要考查代数式化简求值,注意不必按常规用平方差公式和完全平方公式,那样比较复杂,可以直接用提取公因式法化简求值.
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