Dãy trên là một CSN với công bội q = 1/3, số hạng đầu là -1/3
=> Tổng 20 số hạng đầu dãy là: \(\dfrac{-\dfrac{1}{3}\left(\left(\dfrac{1}{3}\right)^{20}-1\right)}{\dfrac{1}{3}-1}\)
B = 1 + \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{20}}\)
Thực hiện phép tính: a) \(11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
b) \(2\dfrac{17}{20}-1\dfrac{11}{15}+6\dfrac{9}{20}:3\) c) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
d) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(A) x +\dfrac{1 }{3 } = \dfrac{2 }{5 }-\dfrac {(-1) }{3} B) \dfrac{5 }{8 } - x = \dfrac{-3 }{20 } - \dfrac{(-1 )}{6 }\)
Rút gọn: \(\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...+\dfrac{1}{3^{19}}-\dfrac{1}{3^{20}}\)
-\(\dfrac{4}{7}\)+\(\dfrac{15}{4}\)-(\(\dfrac{11}{4}\)+\(\dfrac{3}{7}+\dfrac{1}{2}-\dfrac{1}{3}\))
\(\dfrac{1}{5}\left(\dfrac{1}{2}-\dfrac{1}{3}\right):\left(\dfrac{-9}{10}\right)+\dfrac{-7}{3}\)
\(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
giúp mik với
Tìm x, biết:
a) \(\left(3\dfrac{1}{2}-2.x\right).3\dfrac{1}{3}=7\dfrac{1}{3}\)
b) \(\dfrac{4}{9}.x=\dfrac{9}{8}-0,125\)
c) \(\dfrac{-X}{21}=\dfrac{20}{7}\)
B=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)
1)\(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
2)\(\dfrac{ }{\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}}\)
3)\(\dfrac{ }{\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}}\)
4)\(\dfrac{ }{\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}}\)
a) x + \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\)
b) x - \(\dfrac{2}{5}\) = \(\dfrac{2}{7}\)
c) \(\dfrac{3}{5}\) - x = \(\dfrac{1}{10}\)
d) x . \(\dfrac{3}{4}\) = \(\dfrac{9}{20}\)
e) x : \(\dfrac{1}{7}\) = 14
f) ( \(\dfrac{1}{4}\) + x ) . \(\dfrac{1}{2}\) = \(\dfrac{2}{5}\)
g) x . \(\dfrac{2}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{9}{12}\)
h) \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
k) \(3\dfrac{4}{5}\) - x = \(\dfrac{18}{5}\)
l) x . \(2\dfrac{1}{3}\) = \(\dfrac{3}{4}\)
m) x . \(\dfrac{6}{11}\) + x . \(\dfrac{5}{11}\) = 2025
n) x . \(\dfrac{14}{9}\) - x . \(\dfrac{7}{9}\) + x . \(\dfrac{5}{9}\) = 2
Các bạn làm theo cách bình thường ở lớp 5 cho mính nhé!
Chú ý: dấu "." là dấu nhân.
