tính:\(\dfrac{1}{25}+\dfrac{1}{625}+\dfrac{1}{3125}\)
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1,thực hiện phép tính bằng cách thuận tiện nhất
a,1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\)
\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)
\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)
\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)
\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)
\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)
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tinh
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\)
\(=1\)
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\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
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Tính :
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{4}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
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Thực hiện phép tính \(\left(\dfrac{1}{\sqrt{625}}+\dfrac{1}{5}+1\right)\div\left(\dfrac{1}{25}-\dfrac{1}{\sqrt{25}}-1\right)\)
\(\left(\dfrac{1}{\sqrt{625}}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{\sqrt{25}}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\left(-\dfrac{29}{25}\right)\)
\(=\dfrac{31}{25}.\left(-\dfrac{25}{29}\right)\)
\(=-\dfrac{31}{29}\)
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Thuc hien phep tinh;
a/ 1dfrac{4}{23}+dfrac{5}{21}-dfrac{4}{23}+0,5+dfrac{16}{21}
b/ left(dfrac{1}{25}+dfrac{1}{5}+1right):left(dfrac{1}{25}-dfrac{1}{5}-1right)
c/ dfrac{dfrac{1}{9}-dfrac{1}{7}-dfrac{1}{11}}{dfrac{4}{9}-dfrac{4}{7}-dfrac{4}{11}}+ dfrac{0,6-dfrac{3}{25}-dfrac{3}{125}-dfrac{3}{625}}{dfrac{4}{5}-0,16-dfrac{4}{125}-dfrac{4}{625}}
d/ dfrac{2^{12}.3^5-4^6.9^2}{left(2^2.3right)^6+8^4.3^5}-dfrac{5^{10}.7^3-25^5.49^2}{left(125.7right)^3+5^9.14^3}
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Thuc hien phep tinh;
a/ \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
b/ \(\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
c/ \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)+ \(\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)
d/ \(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
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Bài 1 : Tính
a) Adfrac{dfrac{1}{9}-dfrac{1}{7}-dfrac{1}{11}}{dfrac{4}{9}-dfrac{4}{7}-dfrac{4}{11}}+dfrac{0,6-dfrac{3}{25}-dfrac{3}{125}-dfrac{3}{625}}{dfrac{4}{5}-0,16-dfrac{4}{125}-dfrac{4}{625}}
b)Bdfrac{8}{9}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-...-dfrac{1}{6}-dfrac{1}{2}
c)Cdfrac{1}{1.3}-dfrac{1}{2.4}+dfrac{1}{3.5}-dfrac{1}{4.6}+dfrac{1}{5.7}-dfrac{1}{6.8}+dfrac{1}{7.9}-dfrac{1}{8.10}
1 tiếng nữa mình cần rồi giúp mình nhé !!!!!!!!!! Vạn lời tri ân
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Bài 1 : Tính
a) A=\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)\(+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)
b)B=\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
c)C=\(\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
1 tiếng nữa mình cần rồi giúp mình nhé !!!!!!!!!! Vạn lời tri ân
Tính nhanh:
a) A 1 + dfrac{1}{5}+dfrac{1}{25}+dfrac{1}{125}+dfrac{1}{625}+...+dfrac{1}{78125}
b) B dfrac{1}{3}+dfrac{1}{12}+dfrac{1}{48}+dfrac{1}{192}+dfrac{1}{768}+...+dfrac{1}{36864}
c) M dfrac{1}{3}+dfrac{1}{6}+dfrac{1}{12}+dfrac{1}{20}+dfrac{1}{30}+...+dfrac{1}{9900}
d) P dfrac{1}{1+2}+dfrac{1}{1+2+3}+dfrac{1}{1+2+3+4}+...+dfrac{1}{1+2+3+4+...+2018}
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Tính nhanh:
a) A= 1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\)
b) B= \(\dfrac{1}{3}+\dfrac{1}{12}+\dfrac{1}{48}+\dfrac{1}{192}+\dfrac{1}{768}+...+\dfrac{1}{36864}\)
c) M= \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)
d) P= \(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+2018}\)
tính giá trị của biểu thức a) log_5125 và log_6216b) log_{10}dfrac{1}{10000} và logsqrt{1000}c) 81^{log_35} và 125^{log_52}d) left(dfrac{1}{49}right)^{log_7dfrac{1}{8}} và left(dfrac{1}{625}right)^{log_52}
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tính giá trị của biểu thức
a) \(log_5125\) và \(log_6216\)
b) \(log_{10}\dfrac{1}{10000}\) và \(log\sqrt{1000}\)
c) \(81^{log_35}\) và \(125^{log_52}\)
d) \(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}\) và \(\left(\dfrac{1}{625}\right)^{log_52}\)
\(log_5125=log_55^3=3\)
\(log_6216=log_66^3=3\)
\(log_{10}\dfrac{1}{10000}=log_{10}10^{-4}=-4\)
\(log\sqrt{1000}=log_{10}10^{\dfrac{3}{2}}=\dfrac{3}{2}\)
\(81^{log_35}=3^{3log_35}=3^{log_3125}=125\)
\(125^{log_52}=5^{3log_52}=5^{log_58}=8\)
\(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}=7^{-2log_7\dfrac{1}{8}}=7^{log_764}=64\)
\(\left(\dfrac{1}{625}\right)^{log_52}=5^{-4log_52}=5^{log_5\dfrac{1}{16}}=\dfrac{1}{16}\)
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Bài 1: Thực hiện phép tính:
1) \(\dfrac{-17}{30}-\dfrac{11}{-15}+\dfrac{-7}{12}\)
2) \(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
3) \(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}-\dfrac{18}{25}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
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