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(1)化简(0.027)−13−(−17)−2+(279)12−(2−1)0−5iog545;(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+14)+•f(x−14)的定义域.

题目详情
(1)化简(0.027)
1
3
−(−
1
7
)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
(0.027)
1
3
−(−
1
7
)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
)
1
3
−(−
1
7
)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
)
1
3
−(−
1
7
)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
1
3
−(−
1
7
)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
1
3
1133
1
7
1177)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
)−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
−2+(2
7
9
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
7
9
7799)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
)
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
1
2
−(
2
−1)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
1
2
1122
2
2
2)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
)0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
0−5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
5iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
iog545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
g545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
g545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
g545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
g545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
545;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
45;
(2)若函数y=f(x)的定义域为[-1,1],求函数y=f(x+
1
4
)+•f(x−
1
4
)的定义域.

y=f(x+
1
4
)+•f(x−
1
4
)的定义域.
1
4
1144•f(x−
1
4
)的定义域.
1
4
1144
▼优质解答
答案和解析
(1)原式=[(0.3)3]
1
3
−72+(
25
9
)
1
2
−1−45
=
10
3
−49+
5
3
−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
原式=[(0.3)3]
1
3
−72+(
25
9
)
1
2
−1−45
=
10
3
−49+
5
3
−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
3]
1
3
−72+(
25
9
)
1
2
−1−45
=
10
3
−49+
5
3
−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
1
3
111333−72+(
25
9
)
1
2
−1−45
=
10
3
−49+
5
3
−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
2+(
25
9
252525999)
1
2
−1−45
=
10
3
−49+
5
3
−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
1
2
111222−1−45
=
10
3
−49+
5
3
−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
10
3
101010333−49+
5
3
555333−1−45=−90.
(2)
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
3
4
≤x≤
3
4
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
−1≤x+
1
4
≤1
−1≤x−
1
4
≤1
−1≤x+
1
4
≤1−1≤x+
1
4
≤1−1≤x+
1
4
111444≤1−1≤x−
1
4
≤1−1≤x−
1
4
≤1−1≤x−
1
4
111444≤1∴
3
4
≤x≤
3
4
3
4
333444≤x≤
3
4
333444.