Chứng tỏ rằng : 1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
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HH
0
AH
Akai Haruma
Giáo viên
2 tháng 5 2018
Lời giải:
Ta có:
\(\left\{\begin{matrix} \frac{1}{13}< \frac{1}{12}\\ \frac{1}{14}< \frac{1}{12}\\ \frac{1}{15}< \frac{1}{12}\end{matrix}\right.\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}(1)\)
\(\left\{\begin{matrix} \frac{1}{61}< \frac{1}{60}\\ \frac{1}{62}< \frac{1}{60}\\ \frac{1}{63}< \frac{1}{60}\end{matrix}\right.\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}(2)\)
Từ \((1);(2)\Rightarrow \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)
Hay \( \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)
Ta có đpcm.
CT
3 tháng 5 2018
Đặt A là biểu thức đó
Ta có:
\(\dfrac{1}{13}< \dfrac{1}{12};\dfrac{1}{14}< \dfrac{1}{12};\dfrac{1}{15}< \dfrac{1}{12}\)
\(\Rightarrow\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}\)
Ta cũng có
\(\dfrac{1}{61}< \dfrac{1}{60};\dfrac{1}{62}< \dfrac{1}{60};\dfrac{1}{63}< \dfrac{1}{60}\)
\(\Rightarrow\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
\(\Rightarrow\)dpcm
LN
0
NT
2
31 tháng 5 2015
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}
31 tháng 5 2015
Ta có:
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)<1/5+1/12.3+1/60.3
=>S<1/5+1/4+1/20=10/20
Hay S<1/2
LM
0
PT
1
NH
2
HT
20 tháng 6 2015
Ta có:
\(\frac{1}{5}=\frac{1}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}
20 tháng 6 2015
Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
3 tháng 5 2017
Giải:
Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Nhận xét:
\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)
27 tháng 3 2018
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{5}+\dfrac{1}{13}\cdot3+\dfrac{1}{61}\cdot3\\ =\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{3}{60}=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
=> Điều phải chứng minh
TA có:
1/12>1/13
1/12>1/14
1/12>1/15
=>1/12.3=1/4>1/13+1/14+1/15
1/60>1/61
1/60>1/62
1/60>1/63
=>1/60.3=1/20>1/61+1/62+1/63
=>1/5+1/4+1/20> 1/5+1/13+1/14+1/15+1/61+1/62+1/63
=>1/2> 1/5+1/13+1/14+1/15+1/61+1/62+1/63