Đề có vấn đề A= 1/30 +...
1/30 + 1/42 +1/56 +1/72+1/90+1/110+1/132
= 1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12
=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12
= 1/5 -1/12
=7/60
MK lam bai nay roi nen mk nghi de sai !
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{6}-\frac{1}{12}\)
\(A=\frac{1}{12}\)
bài này chứng minh thì bn tự lm típ
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
=>\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
=>\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
=>\(A=\frac{1}{6}-\frac{1}{12}\)
=>\(A=\frac{1}{12}< 1\)
Vậy A<1 (đpcm)
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{6}-\frac{1}{12}=\frac{2}{12}-\frac{1}{12}=\frac{1}{12}\)
Mà : \(\frac{1}{2}< 1\)
Nên : \(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90} +\frac{1}{110}+\frac{1}{132}< 1\)
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{11.12}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{6}-\frac{1}{12}\)
Vì \(\frac{1}{6}\) bé hơn 1 và \(\frac{1}{12}\)lớn hơn 0 nên hiệu 2 số này,tức A sẽ bé hơn 1.
Vậy A<1.
Chúc em học tốt^^
A= \(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}+\frac{1}{10x11}+\frac{1}{11x12}\)<1
A= \(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)<1
A=\(\frac{1}{6}-\frac{1}{12}\)<1
A=\(\frac{1}{12}\)<1
A=\(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
ta thấy \(\frac{1}{6\cdot7}=\frac{1}{6}-\frac{1}{7},\frac{1}{7\cdot8}=\frac{1}{7}-\frac{1}{8}\)
=> A=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}_{ }-\frac{1}{8}......+\frac{1}{11}-\frac{1}{12}\)
=\(\frac{1}{6}-\frac{1}{12}\)
\(=\frac{1}{12}\)<1
\(A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{6}-\frac{1}{12}\)
\(A=\frac{1}{12}< 1\)
