CHỨNG MINH:
\(\frac{ }{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{63}+\frac{1}{64}>3}\)
Những câu hỏi liên quan
Chứng minh :\(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{63}+\frac{1}{64}>4\)
Các bạn xem mình giải có đúng không:
\(1+\frac{1}{2}+........+\frac{1}{63}+\frac{1}{64}\)
\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+...........+\)\(\left(\frac{1}{33}+\frac{1}{34}+...........+\frac{1}{64}\right)\)
\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...........+\left(\frac{1}{64}+.....+\frac{1}{64}\right)\)
=\(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
=4
Vậy \(1+\frac{1}{2}+..............+\frac{1}{64}>4\)
Đúng 0
Bình luận (0)
Chứng minh \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{63}+\frac{1}{64}>4\)
cho:
a) A= 2+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+\frac{1}{65}+\frac{1}{66}+\frac{1}{67}\)
chứng minh rằng A>5
b) B= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{89^2}+\frac{1}{90^2}\)
chứng minh rằng \(\frac{40}{91}\)<B<1
Chứng tỏ rằng :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
A=1+(1/2 + 1/3 + 1/4)+(1/5 + 1/6 + 1/7 + 1/8)+(1/9+...+1/16)+(1/17+...+1/32)+(1/33+...+1/64)
A>1+(1/2 + 1/4 + 1/4)+(1/8+ 1/8+ 1/8+ 1/8)+(1/16+1/16+...+1/16)+(1/64+...+1/64)
A>1 + 1 + 1/2 + 1/2 + 1/2+ 1/2
A>4
Đúng 0
Bình luận (2)
Chung minh rang : 1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}+\frac{1}{64}>4\)
Ta có:\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{64}\)
=\(1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+.........+\left(\frac{1}{33}+......+\frac{1}{64}\right)\)
\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...+\left(\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}\right)\)
=\(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
=4
Vậy \(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{64}>4\)
Đúng 0
Bình luận (0)
Chứng tỏ:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
Chứng tỏ:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}>4\)
Chứng minh rằng H>2
\(H=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.......+\frac{1}{63}\)
Ta có: H=(1/2+1/3+1/4)+(1/5+...+1/8)+(1/9+1/16)+(1/17+...+1/63)
=> H=13/12 + (1/5+...+1/8)+(1/9+...+1/16)+(1/17+...+1/63)
=> H> 1 + 4x(1/8) + 8x (1/16) + (1/17+...+1/63)
=> H> 1+ 1/2 + 1/2 + (1/17+...+1/63)
=> H> 1+1+(1/17+...+1/63)
=> H>1+1
=> H>2
Đúng 0
Bình luận (0)
21 tháng 4 2019 lúc 21:03
a) Chứng minh: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}< \frac{3}{2}\)
b) Chứng minh: \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)