题目内容
化简与求值:
(1)2(2a2+9b)+3(-5a2-4b)
(2)
x-2(x-
y2)+(-
x+
y2),且|x-2|+(y-1)2=0.
(1)2(2a2+9b)+3(-5a2-4b)
(2)
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分析:(1)先去括号,再合并同类项即可;
(2)先去括号,再合并同类项,利用非负数的性质求得x、y的值,再代入即可.
(2)先去括号,再合并同类项,利用非负数的性质求得x、y的值,再代入即可.
解答:解:(1)原式=4a2+18b-15a2-12b
=4a2-15a2-12b+18b
=-11a2+6b;
(2)原式=
x-2x+
y2-
x+
y2
=
x-2x-
x+
y2+
y2
=-3x+y2;
|x-2|+(y-1)2=0
所以x-2=0,y-1=0,
x=2,y=1,
所以原式=-3×2+12
=-6+1
=-5.
=4a2-15a2-12b+18b
=-11a2+6b;
(2)原式=
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=
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=-3x+y2;
|x-2|+(y-1)2=0
所以x-2=0,y-1=0,
x=2,y=1,
所以原式=-3×2+12
=-6+1
=-5.
点评:此题考查合并同类项的方法,非负数的性质,代数式求值等知识点.
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