Tính nhanh:
C=( \(\frac{7}{9}\)+ 1)( \(\frac{7}{20}\)+ 1)( \(\frac{7}{33}\)+ 1).....( \(\frac{7}{10800}\)+ 1).
Tính:
\(C=\left(\frac{7}{9}+1\right)\left(\frac{7}{20}+1\right)\left(\frac{7}{33}+1\right)...\left(\frac{7}{10800}+1\right)\)
Tính nhanh:
C=\(\left(\frac{7}{9}+1\right)\left(\frac{7}{20}+1\right)\left(\frac{7}{33}+1\right).......\left(\frac{7}{10800}+1\right)\)
tính nhanh A= \(\left(\frac{7}{9}+1\right).\left(\frac{7}{20}+1\right).\left(\frac{7}{33}+1\right)....\left(\frac{7}{10800}+1\right)\)
1) Tính:\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
2) Tìm tất cả các số nguyên tố x,y sao cho x2 - 6y2 - 1 = 0
3) Cho \(n\in N\)biết n-10; n+4. n+60 đều là số nguyên tố. CMR: n+90 là số nguyên tố
4) Tính nhanh
\(A=\left(\frac{7}{9}+1\right)\left(\frac{7}{20}+1\right)\left(\frac{7}{33}+1\right).....\left(\frac{7}{10800}+1\right)\)
Các bn giúp mk nhanh lên nhé
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
*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
tính A=\(\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)......\left(1+\frac{7}{2009}\right)\)
Đặt \(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)+...+\left(1+\frac{7}{2009}\right)\)
\(\Leftrightarrow1+\left(\frac{7}{9}.\frac{7}{20}.\frac{7}{33}.\frac{7}{48}.....\frac{7}{2009}\right)\)
Dãy phân số trên có số phân số là:
(2009 - 9) : 4 + 2 =502
\(\Rightarrow A=1+\left(\frac{7^{502}}{9.20.33.48.....2009}\right)\)
A=16/9 .27/20 . 40/33 . 55/48 ....2016/2009
=(16.27.40.55...2016) / (9.20.33.48...2009)
= [(2.8)(3.9)(4.10)(5.11)...(42.48)] / [(1.9)(2.10)(3.11)(4.12)...(41.49)]
=[(2.3.4.5..42)(8.9.10.11..48)] / [(1.2.3.4...41)(9.10.11.12...49)]
=(42.8) / (1.49)
=336/49
=48/7
Cậu thông cảm,mk trình bày hơi khó nhìn
tính tổng\(A=\left(1+\frac{7}{9}\right).\left(1+\frac{7}{20}\right).\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{2900}\right)\)
1.Tính
\(\left(1+\frac{7}{9}\right).\left(1+\frac{7}{20}\right).\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{153}\right).\left(1+\frac{7}{180}\right)\)
Tính
\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)....\left(1+\frac{7}{2900}\right)\)
Cô giải như sau Minh nhé :)
\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{2900}\right)=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{2907}{2900}\)
\(=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}....\frac{57.51}{58.50}=\frac{\left(8.9.10....57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right)\left(2.3.4....50\right)}=\frac{8.51}{58}=\frac{204}{29}\)
( 1 + 7/9 ) x ( 1 + 7/20 ) x ( 1 + 7/33 ) x...x ( 1 + 7/2900)
= (8x2)/(9x1) x (9x3)/(10x2) x (10x4)/(11x3) x...x (57x51)(58x50)
=(8x2x9x3x10x4x...x57x51) / (9x1x10x2x11x3x...x58x50) Sau khi giản ước ta được :
= (8x51) / (1x58) = 204/29