九年级上学期数学计算题专题训练

zhao 初中 2024-04-28 18 1

九年级上学期数学计算题专题训练

回答1

一、有理数的运算（共10小题）
题目:
1. (5) - (-3) + 2(4-6)
2. (-7) * [(8-9)^2 / 4]
3. √[((-3)^3 )/ ((6)^2)]
4. (-5)^2 * (5^2) - (5^-2)
5. ÷[-(2/3), (3/4)] * [-(4/5), (5/6)]
6. (-5)^(-3) * 5^(-2)
7. |-3| + |-7| + |-5|
8. (-1)^2018 + (-1)^2019
9. ∑从i=1到n的i^2
10. ∏从j=1到m的(2j+1)
解答:
1. (5) - (-3) + 2(4-6) = 5 + 3 + 2(-2) = 8 - 4 = 4
2. (-7) * [(8-9)^2 / 4] = (-7) * [(1)^2 / 4] = (-7) * (1 / 4) = -7/4
3. √[((-3)^3 )/ ((6)^2)] = √[(-27)/(36)] = √[-3/-1] = √3
4. (-5)^2 * (5^2) - (5^-2) = (25)(25) - (1/25) = 625 - 1/25 = 624.8
5. ÷[-(2/3), (3/4)] * [-(4/5), (5/6)] = [-8/12, 4/3] * [-4/5, 10/6] = [(-8)/12, 16/9] * [(-4)/5, 10/6] = [-8/15, 160/54] = [-16/30, 160/54] = [-8/15, 80/27]
6. (-5)^(-3) * 5^(-2) = (-5)^(-3) * (1/5)^2 = (1/(25))^(1/2) = 1/sqrt(25) = 1/5
7. |-3| + |-7| + |-5| = |3| + |7| + |5| = 3 + 7 + 5 = 15
8. (-1)^2018 + (-1)^2019 = 1 + (-1) = 1 - 1 = 0
9. ∑从i=1到n的i^2 = 1^2 + 2^2 + ... + n^2 根据等差数列求和公式：Sn = n(a1 + an)/2 和等差数列通项公式：an = a1 + d(n-1) Sn = n((a1 + an)/2) = n(a1 + an)/2 代入公式：Sn = n(1 + n^2)/2 将n替换为n^2：S_n^2 = n^3(1 + n^2)/2 所以：∑从i=1到n的i^2 = n(n+1)(2n+1)/6
10. ∏从j=1到m的(2j+1) = (21+1)(22+1)...(2m+1) = 35...*(2m+1) 这是首项为3，公比为5的等比数列的前m项乘积，其结果为：(5^m - 3^m) / (5 - 3) = (5^m - 3^m) / 2
答案:
1. 4
2. -7/4
3. √3
4. 624.8
5. [-8/15, 80/27]
6. 1/5
7. 15
8. 0
9. n(n+1)(2n+1)/6
10. (5^m - 3^m) / 2
二、实数的指数和对数运算（共5小题）
题目:
1. log(x^y)以2为底
2. e^ln(e^2)
3. ln(10^3)
4. (log_b c)^2 - (log_c b)^2
5. (10^x)^(log_10 x)
解答:
1. log(x^y)以2为底 = y * log_2 x
2. e^ln(e^2) = e^2
3. ln(10^3) = 3 * ln(10)
4. (log_b c)^2 - (log_c b)^2 = (log_bc)^2 - (log_cb)^2 = (log_ec)^2 - (log_ce)^2 = (c^e)^2 - (e^c)^2 = ce^c - ec^e
5. (10^x)^(log_10 x) = 10^(x * log_10 x) = 10^(log_10 x^x) = x^x
答案:
1. y * log_2 x
2. e^2
3. 3 * ln(10)
4. ce^c - ec^e
5. x^x
zhao 2024-04-28 0 回复

回答1

一、有理数的运算（共10小题）

二、实数的指数和对数运算（共5小题）