Chứng tỏ rằng: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}<2\)
Chứng tỏ rằng: \(1< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+......+\frac{1}{16}+\frac{1}{17}< 2\)2
Chứng tỏ rằng: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}
1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 < 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 6/5 (1)
1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 < 1/11 + 1/11 + 1/11 + 1/11 +1/11 + 1/11 + 1/11 = 7/11 (2)
Từ (1) và (2) => :
A < 6/5 + 7/11 = 101/55 < 110/55 = 2
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Chứng Tỏ Rằng\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}<2\)
Chứng tỏ rằng
C = \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...........+\frac{1}{16}+\frac{1}{17}<2\)
ĐANG CẦN GẤP
ai trả lời nanh , đúng cho like
C=1/5+1/6+1/7+..........+1/16+1/17
C=(1/5+1/6+........+1/9)+(1/10+1/11+........+1/17)
CÓ :1/5>1/6>1/7>1/8>1/9 => 1/5+1/6+........+1/9 < 1/5.5=1
1/10>1/11>1/12>....>1/16>1/17 => 1/10+1/11+........+1/17 < 1/10.8=4/5
=> C<1+4/5 => C<9/5 MÀ 9/5<2 NÊN C<2(ĐPCM)
Chứng tỏ rằng
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}<2\)
Chứng tỏ rằng : \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}<2\)
Đặt A = 1/5+1/6+1/7+...+1/17
Ta có :
1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 < 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 6/5 (1)
1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 < 1/11 + 1/11 + 1/11 + 1/11 +1/11 + 1/11 + 1/11 = 7/11 (2)
Từ (1) và (2) => :
A < 6/5 + 7/11 = 101/55 < 110/55 = 2
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)
\(=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+...+\frac{1}{17}\right)
Ta có:
1/5+1/6+1/7+1/8+1/9<1/5.5=1 (*)
1/10+1/11+....+1/16+1/17<1/8.8=1 (**)
Cộng (*) với (**) ta được:
1/5+1/6+1/7+....+1/17<2(đpcm)
Chứng tỏ rằng\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}<2\)
chứng tỏ\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 2\)
Đặt A=\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)\(=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}\right)\)
Có: \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}< \frac{1}{5}.6=\frac{6}{5}\)(1)
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}< \frac{1}{11}.7=\frac{7}{11}\)(2)
Từ (1) và (2) suy ra: A\(< \frac{6}{5}+\frac{7}{11}=\frac{101}{55}\)
Lại có: \(\frac{101}{55}< \frac{110}{55}=2\)
Suy ra: A<2 (đpcm)
Chứng minh rằng
\(\frac{1}{5}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{44}+\frac{1}{45}>\frac{5}{6}\)