Các câu hỏi tương tự
1.Tính nhanh
a)427-98
b)2*19*15+3*43*10+62*80
c)\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
d)\(\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{2}{9}\right)\cdot\left(1-\frac{3}{90}\right)\cdot.........\cdot\left(1-\frac{2018}{9}\right)\)
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
Tính nhanh na
Tính
a) \(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\frac{2}{243}+\frac{2}{729}\)
b) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
Tìm a
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right).\)
\(\left[x+\frac{1}{3}\right]+\left[x+\frac{1}{15}\right]+\left[x+\frac{1}{35}\right]+...+\left[x+\frac{1}{575}\right]\)
\(=11xx+\left[\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right]\)
Chứng minh rằng:frac{1}{5}+frac{1}{15}+frac{1}{25}+....+frac{1}{1985} frac{9}{20}mk làm thế này đúng ko mọi ngườiĐặt Afrac{1}{3}+frac{1}{5}+frac{1}{7}+frac{1}{9}+......+frac{1}{243}Afrac{1}{3}+left(frac{1}{5}+frac{1}{7}+frac{1}{9}right)+left(frac{1}{11}+frac{1}{13}+frac{1}{15}+....+frac{1}{27}right)+left(frac{1}{29}+frac{1}{31}+frac{1}{33}+....+frac{1}{81}right)+left(frac{1}{83}+frac{1}{85}+frac{1}{87}+.....+frac{1}{243}right)Afrac{1}{3}+frac{1}{9}.3+frac{1}{27}.9+frac{1}{81}.27+frac{1}{243}.81f...
Đọc tiếp
Chứng minh rằng:\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+....+\frac{1}{1985}< \frac{9}{20}\)
mk làm thế này đúng ko mọi người
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+......+\frac{1}{243}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+....+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+....+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+.....+\frac{1}{243}\right)\)
\(=>A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81=\frac{1}{3.5}=\frac{5}{3}\)
\(=>A>\frac{5}{3}>\frac{5}{4}=>A< \frac{5}{4}\)
\(=>\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)
\(=>1+\frac{1}{3}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)
\(=>\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}\right)< \frac{9}{4}.\frac{1}{5}\)
\(=>\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+......+\frac{1}{1985}< \frac{9}{20}\)
\(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\frac{2}{243}+\frac{2}{729}\)
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+...+50}\)
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729}\)