\(E=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\\ 3E=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\\ 3E-E=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\\ 2E=1-\frac{1}{3^8}\\ E=\frac{1-\frac{1}{3^8}}{2}\)
a) \(\frac{2}{5}+\left(-\frac{4}{3}\right)+\left(-\frac{1}{2}\right)\)
b) \(\frac{5}{8}-\left(-\frac{2}{5}-\frac{3}{10}\right)\)
c) \(\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\left(\frac{1}{4}+\frac{3}{8}\right)\right]\)
d) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
e) \(\left(-\frac{1}{3}+\frac{5}{6}\right).11-7\)
f) \(\left(-\frac{5}{9}\right).\frac{3}{11}+\left(-\frac{13}{18}\right).\frac{3}{11}\)
bài 2:tính tổng đặc biệt:
\(E=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
Bài 3:chứng minh:
a,\(A=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}\)chứng minh rằng \(A⋮100\)
b,\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)chứng minh rằng \(A>\frac{4}{3}\)
hlep me!!!!
Cho A= \(^{\frac{2^{2018}}{2^{2018}+3^{2019}}}+\frac{3^{2019}}{3^{2019}+5^{2020}}+\frac{5^{2020}}{5^{2020}+2^{2018}}\)
Và B=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)
So sánh A và B
A= \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(\frac{1}{4}+\frac{1}{3}\left(2x-1\right)=-5\)
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(5\frac{4}{7}:x=13\)
\(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
làm hộ mik nhanh vs ... mik chỉ còn 50p nữa là nộp pài r
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
Tính: \(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\)
bài 1:chứng minh.
\(a,A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}.\)chứng minh rằng \(A⋮100\)
\(b,A=\frac{1}{11}+\frac{1}{12}+\frac{2}{13}+...+\frac{1}{70}.\)chứng minh răng \(A>\frac{4}{3}\)
ai đó giúp với.
tìm x biết :
\(a,\frac{1}{2}-\frac{2}{3}x=\frac{7}{12}\) b,/x+\(\frac{1}{2}\)/=\(\frac{3}{4}\)
