试题

题目:
化简或求值(本题满分16分,5+5+6):
(1)2x2-2+七x-1-2x-x2;           
(2)a2-(七a2-b2)-七(a2-2b2
(七)已知:(x-七)2+|y+2|=0,求代数式2x2+(-x2-2xy+2y2)-2(x2-xy+2y2)的值.
答案
解:(1)2x2-2+3x-1-2x-x2
=x2+x-3;
(2)原式=a2-3a2+b2-3a2+6b2
=-5a2+7b2
(3)∵(x-3)2+|y+2|=6,
∴x=3,y=-2,
2x2+(-x2-2xy+2y2)-2(x2-xy+2y2
=2x2-x2-2xy+2y2-2x2+2xy-4y2
=-x2-2y2
当x=3,y=-2时,原式=-17.
解:(1)2x2-2+3x-1-2x-x2
=x2+x-3;
(2)原式=a2-3a2+b2-3a2+6b2
=-5a2+7b2
(3)∵(x-3)2+|y+2|=6,
∴x=3,y=-2,
2x2+(-x2-2xy+2y2)-2(x2-xy+2y2
=2x2-x2-2xy+2y2-2x2+2xy-4y2
=-x2-2y2
当x=3,y=-2时,原式=-17.
考点梳理
整式的加减—化简求值;非负数的性质:绝对值;非负数的性质:偶次方;合并同类项;去括号与添括号.
(1)直接合并同类项即可求解;
(2)首先去括号,然后合并同类项即可求解;
(3)首先根据非负数的性质求出x、y的值,然后把所给的多项式化简,最后代入已知数据计算即可求解.
此题主要考查了整式的化简求值,同时也利用了非负数的性质,解题时首先去括号、合并同类项,然后代入已知数据计算即可求解.
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