试题

题目:
合并同类项:
(1)
1
6
x-(-
1
3
x)-1+(-
1
2
x)+(-1)

(2)3x2y-4xy2-3+5x2y+2xy2+5;
(3)5x2-[3x-2(2x-3)-4x2];
(4)3(2x2-y2)-2(3y2-2x2)-{-[-(-x2)-y2]}.
答案
解:(1)原式=
1
6
x+
1
3
x-1-
1
2
x-1
=(
1
6
+
1
3
-
1
2
)x+(-1-1)
=-2;
(2)原式=(3+5)x2y+(-4+2)xy2+(-3+5)=8x2y-2xy2+2;
(3)原式=5x2-3x+4x-6+4x2=9x2+x-6;
(4)原式=6x2-3y2-6y2+4x2+x2-y2=11x2-10y2
解:(1)原式=
1
6
x+
1
3
x-1-
1
2
x-1
=(
1
6
+
1
3
-
1
2
)x+(-1-1)
=-2;
(2)原式=(3+5)x2y+(-4+2)xy2+(-3+5)=8x2y-2xy2+2;
(3)原式=5x2-3x+4x-6+4x2=9x2+x-6;
(4)原式=6x2-3y2-6y2+4x2+x2-y2=11x2-10y2
考点梳理
合并同类项.
(1)先简化符号,再合并同类项;
(2)直接根据合并同类项的法则计算;
(3)、(4)先去括号,再合并同类项.
本题考查整式的运算,有括号的要先去括号,再合并同类项.
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