tính:\(\dfrac{1}{25}+\dfrac{1}{625}+\dfrac{1}{3125}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)
\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)
\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)
\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)
\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\)
\(=1\)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
\(\left(\dfrac{1}{\sqrt{625}}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{\sqrt{25}}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\left(-\dfrac{29}{25}\right)\)
\(=\dfrac{31}{25}.\left(-\dfrac{25}{29}\right)\)
\(=-\dfrac{31}{29}\)
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
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
\(\dfrac{1}{25}\)+\(\dfrac{1}{625}\)+\(\dfrac{1}{3125}\)=\(\dfrac{125}{3125}\)+\(\dfrac{5}{3125}\)+\(\dfrac{1}{3125}\)
=\(\dfrac{131}{3125}\)
chúc bạn học tốt ạ