Các câu hỏi tương tự
Bài 1:Tìm số tự nhiên ab sao cho ab -ba là số chính phương
Bài 2:Chứng minh rằng :\(\frac{19}{20!}+\frac{19}{21!}+\frac{19}{22!}+.........+\frac{19}{5000!}<\frac{1}{19!}\)
Giair giùm mình chi tiết nha .Bạn nào đúng sẽ có 2 l-i-k-e
CHO
S=\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{199}+\frac{1}{200}\)
CHỨNG MINH RẰNG S>\(\frac{9}{10}\)
Chứng minh rằng :
\(\frac{7}{12}< \frac{1}{21}+\frac{1}{20}+...+\frac{1}{40}< 1\)
Chú ý p/s thứ 2 là 1/20 chứ k phải 1/22 nha
Chứng minh rằng: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
cho A = \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{58}+\frac{1}{59}\)chứng minh A <\(\frac{3}{2}\)
Chứng minh rằng :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
C/m ::
\(S=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}>\frac{1}{2}\)
1 . Tinh : a , left(1-frac{1}{6}right)left(1-frac{1}{10}right)left(1-frac{1}{14}right).....left(1-frac{1}{5050}right)b,frac{^{2^{19}}.left(3^3right)^3+3.5.left(2^2right)^9.left(3^2right)^4}{left(2.3right)^9.2^{10}+left(3.2^2right)^{10}}c,frac{18.frac{19}{2}.frac{20}{3}.frac{21}{4}.....frac{36}{19}}{20.frac{21}{2}.frac{22}{3}.....frac{36}{17}}giup mjk nha mjk tjk cho
Đọc tiếp
1 . Tinh : a , \(\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{14}\right).....\left(1-\frac{1}{5050}\right)\)b,\(\frac{^{2^{19}}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}\)c,\(\frac{18.\frac{19}{2}.\frac{20}{3}.\frac{21}{4}.....\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}.....\frac{36}{17}}\)giup mjk nha mjk tjk cho
1)
\(Cho:\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{200}\)
Chứng minh: \(A>\frac{9}{10}\)
2)
Cho \(B=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
Chứng minh \(B>\frac{7}{12}\)