题目内容
先化简,再求值:(1)-2(x2+1)+5(x-5)-
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(2)已知(x-
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分析:(1)去括号,合并同类项,再代入求值;
(2)由非负数的性质求得x,y的值,把代数式去括号,合并同类项,再代入求值.
(2)由非负数的性质求得x,y的值,把代数式去括号,合并同类项,再代入求值.
解答:解:(1)原式=-2x2-2+5x-25-2x2+x=(-2x2-2x2)+(5x+x)+(-25-2)=-4x2+6x-27,
将x=-1
代入上式得,
原式=-4×(-1
)2+6×(-1
)-27=-45;
(2)由题意x=
,y=-
,
∴原式=x2-2y2+3xy-x2+3y2-
xy
=(x2-x2)+(-2y2+3y2)+(3xy-
xy)
=y2+
xy=(-
)2+
×
×(-
)
=
-
=-
.
将x=-1
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原式=-4×(-1
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(2)由题意x=
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∴原式=x2-2y2+3xy-x2+3y2-
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=(x2-x2)+(-2y2+3y2)+(3xy-
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=y2+
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| 5 |
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=
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=-
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点评:(1)解决此类题目的关键是熟记去括号法则,熟练运用合并同类项的法则;
(2)非负数的性质:若两个非负数的和为0,则这两个数均为0.
(2)非负数的性质:若两个非负数的和为0,则这两个数均为0.
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