*Cuộc thi toán nâng cao cấp THCS (dành riêng cho khối 6) (vòng 1)
Bài toán 1:
Tính \(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)
Bài toán 2:
Cho \(A=\frac{1}{2^3+3}+\frac{1}{3^3+4}+\frac{1}{4^3+5}+...+\frac{1}{2018^3+2019}\)
Hãy so sánh A với \(\frac{1}{6}\)
XD: best tiếng anh chuyển sang toán ak!?
\(B1:\)
\(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)
\(=\frac{16}{9}\cdot\frac{27}{20}\cdot\frac{40}{33}\cdot\cdot\cdot\frac{10807}{10800}\)
\(=\frac{8.2}{9.1}\cdot\frac{9.3}{10.2}\cdot\frac{10.4}{11.3}\cdot\cdot\cdot\frac{57.51}{58.50}\)
\(=\frac{\left(8.9.10...57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right).\left(1.2.3...50\right)}\)
\(=\frac{8.51}{58.1}=\frac{204}{29}\)
Vậy.....
\(M=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{10800}\right)\)
\(M=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{10807}{10800}\)
\(M=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}...\frac{107.101}{108.100}\)
\(M=\frac{\left(2.3.4...101\right)\left(8.9.10...107\right)}{\left(1.2.3...100\right)\left(9.10.11...108\right)}\)
\(M=\frac{101.8}{108}\)
\(M=\frac{202}{27}\)
k mình nha . câu 2 tí nữa mình gửi
câu 2 đây nha!
Ta có : \(A=\frac{1}{2^3+3}+\frac{1}{3^3+4}+\frac{1}{4^3+5}+...+\frac{1}{2018^3+2019}\)
< \(B=\frac{1}{2^3+3}+\frac{1}{3^3+4}+\frac{1}{4^3}+\frac{1}{5^3}+...+\frac{1}{2018^3}\)
< \(C=\frac{1}{2^3+3}+\frac{1}{3^3+4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{2017.2018.2019}\)
\(C=\frac{1}{11}+\frac{1}{31}+\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(C=\frac{1}{11}+\frac{1}{31}+\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{2018.2019}\right)\)
\(C=\frac{1}{11}+\frac{1}{31}+\frac{1}{24}-\frac{1}{2.2018.2019}\)
Lại có : \(\frac{1}{11}+\frac{1}{31}+\frac{1}{24}< \frac{1}{6}\)
=> C<\(\frac{1}{6}\)
=> A < \(\frac{1}{6}\)
