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Trả lời
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
\(\Leftrightarrow\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)
\(\Leftrightarrow\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}\)
\(\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)< \frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\left(đpcm\right)\)
1/2 lớn hơn
vì phân số 1/2 có mẫu số nhỏ hơn các phân số kia nên phân số 1/2 sẽ lớn hơn các phân số kia
\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{8}\right)-\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}<\frac{1}{2}\)
Ta có: Gọi dãy số cần chứng minh là A
\(A<\frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)\)
\(A<\frac{1}{3}+\frac{3}{30}+\frac{4}{60}\)
\(A<\frac{10}{30}+\frac{3}{30}+\frac{2}{30}\)
\(A<\frac{15}{30}=\frac{1}{2}\)
Vậy \(A<\frac{1}{2}\)
k nha
\(T=\left(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)
\(T
Ta thấy: \(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}\)
\(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\)
\(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}\)
\(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)
Do đó: \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}\cdot3+\frac{1}{45}\cdot3=\frac{1}{2}\)
Đặt A = 1/3 + 1/31 + 1/35 + 1/37 + 1/53 + 1/61
A < 1/3+ ( 1/30+1/30+1/30)+( 1/45+1/45+1/45)
A < 1/3+1/10+1/15
A < 1/2
Chứng tỏ 1/3+1/31+1/35+1/37+1/53+1/61<1/2
k nhé, ủng hộ k, mk trả lời đầu tiên đó
Nhận xét:
\(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{10}\)
\(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}+\frac{1}{45}+\frac{1}{45}=\frac{1}{15}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\) (Đpcm)
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 7/14
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 <1/14 +1/14 +1/14 +1/14 +1/14 +1/14 +1/14
dù 1/3>1/14 nhưng :1/30<1/14 1/32<1/14 ;1/35<1/14 ;1/45<1/14 ;1/47<1/14 ;1/50<1/14
nên: 1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 1/2
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 7/14
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 <1/14 +1/14 +1/14 +1/14 +1/14 +1/14 +1/14
dù 1/3>1/14 nhưng :1/30<1/14 1/32<1/14 ;1/35<1/14 ;1/45<1/14 ;1/47<1/14 ;1/50<1/14
nên: 1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 1/2