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计算(1)(13−521+314−27)÷(−142);(2)(−1)2014÷(−52)×(−53)+|0.8−1|;(3)(-2x2+3x)-[5x-(2x2+1)-x2];(4)−12a−2(a−12b2)−(32a−13b2).
题目详情
计算
(1)(
−
+
−
)÷(−
);
(2)(−1)2014÷(−52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).
(
−
+
−
)÷(−
);
(2)(−1)2014÷(−52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).
1 1 3 3
5 5 21 21
3 3 14 14
2 2 7 7
1 1 42 42
(−1)2014÷(−52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).)2014÷(−52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).)2014÷(−52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).2014÷(−52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).52)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).2)×(−
)+|0.8−1|;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
a−2(a−
b2)−(
a−
b2).
5 5 3 3
222
−
a−2(a−
b2)−(
a−
b2).
1 1 2 2
1 1 2 2 b2)−(
a−
b2).b2)−(
a−
b2).2)−(
a−
b2).
3 3 2 2
1 1 3 3 b2).b2).2).
(1)(
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
1 |
42 |
(2)(−1)2014÷(−52)×(−
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
(
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
1 |
42 |
(2)(−1)2014÷(−52)×(−
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
1 |
42 |
(−1)2014÷(−52)×(−
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
5 |
3 |
222
−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
3 |
2 |
1 |
3 |
3 |
2 |
1 |
3 |
3 |
2 |
1 |
3 |
▼优质解答
答案和解析
解(1)(
−
+
−
)÷(−
);
=(
−
+
−
)×(−42)
=-14+10-9+12
=-1;
(2(−1)2014÷(−52)×(−
)+|0.8−1|;
=1÷(−25)×(−
)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. (
1 1 13 3 3−
5 5 521 21 21+
3 3 314 14 14−
2 2 27 7 7)÷(−
1 1 142 42 42);
=(
−
+
−
)×(−42)
=-14+10-9+12
=-1;
(2(−1)2014÷(−52)×(−
)+|0.8−1|;
=1÷(−25)×(−
)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. (
1 1 13 3 3−
5 5 521 21 21+
3 3 314 14 14−
2 2 27 7 7)×(−42)
=-14+10-9+12
=-1;
(2(−1)2014÷(−52)×(−
)+|0.8−1|;
=1÷(−25)×(−
)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. (−1)2014÷(−52)×(−
)+|0.8−1|;
=1÷(−25)×(−
)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. 2014÷(−52)×(−
)+|0.8−1|;
=1÷(−25)×(−
)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. 2)×(−
5 5 53 3 3)+|0.8−1|;
=1÷(−25)×(−
)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. 1÷(−25)×(−
5 5 53 3 3)+0.2
=
+
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2.
1 1 115 15 15+
1 1 15 5 5
=
;
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2.
4 4 415 15 15;
(3)(-2x22+3x)-[5x-(2x22+1)-x22];
=-2x22+3x-[5x-2x22-1-x22]
=-2x22+3x-5x+3x22+1
=x22-2x+1;
(4)−
a−2(a−
b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. −
1 1 12 2 2a−2(a−
1 1 12 2 2b2)−(
a−
b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. 2)−(
3 3 32 2 2a−
1 1 13 3 3b2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. 2)
=−
a−2a+b2−
a+
b2
=−4a+
b2. −
1 1 12 2 2a−2a+b2−
a+
b2
=−4a+
b2. 2−
3 3 32 2 2a+
1 1 13 3 3b2
=−4a+
b2. 2
=−4a+
b2. −4a+
4 4 43 3 3b2. 2.
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
1 |
42 |
=(
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
=-14+10-9+12
=-1;
(2(−1)2014÷(−52)×(−
5 |
3 |
=1÷(−25)×(−
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
1 |
42 |
=(
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
=-14+10-9+12
=-1;
(2(−1)2014÷(−52)×(−
5 |
3 |
=1÷(−25)×(−
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
1 |
3 |
5 |
21 |
3 |
14 |
2 |
7 |
=-14+10-9+12
=-1;
(2(−1)2014÷(−52)×(−
5 |
3 |
=1÷(−25)×(−
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
5 |
3 |
=1÷(−25)×(−
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
5 |
3 |
=1÷(−25)×(−
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
5 |
3 |
=1÷(−25)×(−
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
5 |
3 |
=
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
1 |
15 |
1 |
5 |
=
4 |
15 |
(3)(-2x2+3x)-[5x-(2x2+1)-x2];
=-2x2+3x-[5x-2x2-1-x2]
=-2x2+3x-5x+3x2+1
=x2-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
4 |
15 |
(3)(-2x22+3x)-[5x-(2x22+1)-x22];
=-2x22+3x-[5x-2x22-1-x22]
=-2x22+3x-5x+3x22+1
=x22-2x+1;
(4)−
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
1 |
2 |
1 |
2 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
3 |
2 |
1 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
=−
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
1 |
2 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
3 |
2 |
1 |
3 |
=−4a+
4 |
3 |
=−4a+
4 |
3 |
4 |
3 |
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