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\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{990}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{6}{25}\)
1/20 + 1/30 + 1/42 + 1/56 +...+ 1/990 = 1/4.5 +1/5.6 +1/6.7 + 1/7.8 +...+ 1/99.100 =1/4 - 1/5 + 1/5 -1/6 + 1/6 -1/7 + 1/7 - 1/8 + ...+1/99 -1/100 =1/4-1/100 = 24/100=6/25
=1/4.5+1/5.6+1/6.8+1/7.8+....1/33.30
=1/4-1/5+1/5-1/6+1/6-1/8+1/7-1/8+...+1/30-1/33
=1/4-1/33
=29/132
\(B1\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)
\(=1-\frac{1}{39}\)
\(=\frac{38}{39}\)
\(B2\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}\)
\(=\frac{24}{100}\)
\(=\frac{6}{25}\)
Bài 1 :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{370}{741}\)
1/20 + 1/30 + 1/42 + ... + 1/9900
= 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/99.100
= 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/99 - 1/100
= 1/4 - 1/100
= 6/25
bn ơi,hình nhưa sai đề,số 990 mik ko phân tích ra tích của 2 số tự nhiên liên tiếp được,chắc là sai đề nha bn,bn kiểm tra lại đề rồi đăng câu hỏi khác nhé!!!,để mik xem lại coi.....
kiểm tra nha,nếu mà đúng đề thì để mik xem lại...
T.T
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/9
=1-1/9
=8/9
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
= \(1-\frac{1}{9}\)
= \(\frac{8}{9}\)
Trần Thùy Dung nó đã bảo \(990\ne99\cdot100\) rùi mà vẫn tách như v
=\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{4}-\frac{1}{100}\)
=\(\frac{24}{100}=\frac{6}{25}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}\)
\(A=\frac{3}{8}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
mình nhé!
2 2/9 - x = 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
20/9 - x = 1/3×4 + 1/4×5 + 1/5×6 + 1/6×7 + 1/7×8 + 1/8×9
20/9 - x = 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/8 - 1/9
20/9 - x = 1/3 - 1/9
20/9 - x = 3/9 - 1/9
20/9 - x = 2/9
x = 20/9 - 2/9
x = 18/9 = 2
Vậy x = 2
Đề thiếu?
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{6}{25}\)