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1/5+1/13+1/25+1/41+1/61+1/85+1/113
=1/5+(1/13+1/25+1/41)+(1/85+1/61+1/113)<15+1/12+1/12+1/12+1/60+1/60+1/60
..............................................................<1/5+1/4+1/20
..............................................................<4/20+5/20+1/20
..............................................................<1/2
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{25}+\frac{1}{41}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)\)
< \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{4}{20}+\frac{5}{20}+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(đpcm)
ê cho hỏi tại sao lại ra < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
Hơi nhầm xíu 113 . 7^2+8^2=113 cứ tưởng 112. Hơi ngáo tí =[[
Lời giải
Biến đổi tương đương ta được: \(L=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}=\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+\dfrac{1}{4^2+5^2}+\dfrac{1}{5^2+6^2}+\dfrac{1}{6^2+7^2}+\dfrac{1}{7^2+8^2}\)
\(L=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+...+\dfrac{1}{7^2+\left(7+1\right)^2}\)
Chứng minh 1 bđt cơ bản sau: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\) thật vậy:
\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1=2n\left(n+1\right)+1>2n\left(n+1\right)\)
\(\Rightarrow\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2n\left(n+1\right)}\)
trở lại bài toán ta có: \(L< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+...+\dfrac{1}{2.7.8}\)
\(L< \dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..+\dfrac{1}{7.8}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{8}\right)=\dfrac{1}{2}-\dfrac{1}{16}< \dfrac{1}{2}\left(đpcm\right)\)
Đề sai đúng hk? CHỗ kia 112 chứ lấy đâu ra 113
p/s : 7^2+8^2=112. =))
đặt A=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}\)
= \(\dfrac{1}{5}+(\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41})+(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113})\)
=> A< \(\dfrac{1}{5}+(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12})+(\dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60})\)
A<\(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)=\(\dfrac{1}{2}\)
vậy A<\(\dfrac{1}{2}\),<2=> A<2 (đpcm)