试题
题目:
按下列条件求代数式(a+b)(a
2
-ab+b
2
)与a
3
+b
3
的值,并根据计算的结果写出你发现的结论.
(1)a=3,b=1; (2)a=2
1
2
,b=1
1
2
.
答案
解:(1)当a=3,b=1时,
(a+b)(a
2
-ab+b
2
)
=(3+1)(3
2
-3×1+1
2
)
=4×7
=28,
a
3
+b
3
=3
3
+1
3
=27+1
=28;
即(a+b)(a
2
-ab+b
2
)=a
3
+b
3
.
(2)当a=2
1
2
=
5
2
,b=1
1
2
=
3
2
时,
(a+b)(a
2
-ab+b
2
)
=(
5
2
+
3
2
)[
(
5
2
)
2
-
5
2
×
3
2
+
(
3
2
)
2
]
=4×
19
4
=19,
a
3
+b
3
=
(2
1
2
)
3
+
(1
1
2
)
3
=
125
8
+
27
8
=19,
结论是(a+b)(a
2
-ab+b
2
)=a
3
+b
3
.
解:(1)当a=3,b=1时,
(a+b)(a
2
-ab+b
2
)
=(3+1)(3
2
-3×1+1
2
)
=4×7
=28,
a
3
+b
3
=3
3
+1
3
=27+1
=28;
即(a+b)(a
2
-ab+b
2
)=a
3
+b
3
.
(2)当a=2
1
2
=
5
2
,b=1
1
2
=
3
2
时,
(a+b)(a
2
-ab+b
2
)
=(
5
2
+
3
2
)[
(
5
2
)
2
-
5
2
×
3
2
+
(
3
2
)
2
]
=4×
19
4
=19,
a
3
+b
3
=
(2
1
2
)
3
+
(1
1
2
)
3
=
125
8
+
27
8
=19,
结论是(a+b)(a
2
-ab+b
2
)=a
3
+b
3
.
考点梳理
考点
分析
点评
代数式求值.
(1)把a=3,b=1分别代入(a+b)(a
2
-ab+b
2
)与a
3
+b
3
求出即可;
(2)把a=2
1
2
,b=1
1
2
分别代入(a+b)(a
2
-ab+b
2
)与a
3
+b
3
求出即可;根据求出的结果即可得出结论.
本题考查了求代数式的值,关键是能根据求出的结果得出结论.
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