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a) \(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2^3\cdot5\cdot10\cdot7}{2^3\cdot5\cdot7\cdot77}=\dfrac{10}{77}\)
\(\dfrac{2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}=\dfrac{2^3\cdot3\cdot5^3\cdot7\cdot3^2\cdot8}{3\cdot2^3\cdot2\cdot5^3\cdot14}=\dfrac{7\cdot3^2\cdot8}{2\cdot14}=\dfrac{63\cdot8}{2\cdot14}=18=\dfrac{1386}{77}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{101.103}\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{101.103}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)
=\(\frac{1}{2}.\frac{102}{103}\)
=\(\frac{51}{103}\)
A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{101.103}\)
A = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{101.103}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{101}\) - \(\dfrac{1}{103}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{103}\))
A = \(\dfrac{1}{2}\). \(\dfrac{102}{103}\)
A = \(\dfrac{51}{103}\)
Em ơi thừa số thứ ba phải là \(\dfrac{1}{5.7}\) mới đúng em nhé.
Ta có :
\(B=\frac{3^2.5^2.7^2.3^3.7^2}{3^2.5^2.7^4}=\frac{3^4.5^2.7^4}{3^2.5^2.7^4}=\frac{3^4}{3^2}=3^2=9\)
Vậy B = 9.
k nha bạn
B=\(\frac{\left(9.25.49\right).\left(9.49\right)}{735^2}\)
B=\(\frac{11025.441}{540225}\)
B=\(\frac{4862025}{540225}\)
B=9
\(B=\frac{\left(3^2.5^2.7^2\right).\left(3^3.7^2\right)}{3.5.7^2}=\frac{3^5.5^2.7^4}{3.5.7^2}=\frac{3^4.5.7^2}{1}=19845\)
\(B=\frac{3^2\times5^2\times7^2\times3^3\times7^2}{3\times5\times7^2}=\frac{3^5\times5^2\times7^4}{3\times5\times7^2}=3^4\times5\times7^2=19845\)
Ta có:
\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
\(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}=\frac{22}{35}\)
(chỉnh đề)
A=\(-1+2-3-4-5+6-7-8-9+...-2021-2022+2023-2024\)
=\(\left(-1-2024\right)+\left(2+2023\right)+\left(-3-2022\right)+\left(-4-2021\right)+\left(-5-2020\right)+\left(6+2019\right)-\left(-7-2018\right)+\left(-8-2017\right)+\left(-9-2016\right)+...+\left(1010+1015\right)+\left(-1011-1014\right)+\left(-1012-1013\right)\)=\(-2025+2025-2025-2025-2025+2025-2025-2025-2025+...+2025-2025-2025\)=253.2025-1771.2025=-3 073 950.
B=\(1.3.5+3.5.7+5.7.9+7.9.11+...+99.101.103\)
8B=\(1.3.5.8+3.5.7.8+5.7.9.8+7.9.11.8+...+99.101.103.8\)
8B=\(1.3.5.\left[7-\left(-1\right)\right]+3.5.7.\left(9-1\right)+5.7.9.\left(11-3\right)+7.9.11.\left(13-5\right)+...+99.101.103.\left(105-97\right)\)8B=\(3.5+3.5.7+3.5.7.9-3.5.7+5.7.9.11-3.5.7.9+7.9.11.13-5.7.9.11+...+99.101.103.105-97.99.101.103\)
B=\(\dfrac{3.5+99.101.103.105}{8}=13517400\)
3.5.7/3.5 + 3.7
Vì 3.5 chia cho 3.5 là hết nên chỉ còn lại
7 + 3.7 = 7. (3+1) = 7.4 = 28