=> 0.4614154846 < 0.5

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

không biết ?

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

Ta có:\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)

\(=\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)\(< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)\(=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)

Vậy ............

Ta có: 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/3 + 3/31 + 3/47 < 1/3 + 3/30 + 3/45

= 1/3 + 1/10 + 1/15 = 1/3 + (1/30) * (3+2) = 1/3 + (1/0) * 5 = 1/3 + 1/6

= (1/6) * (2+1) = (1/6) * 3 = 1/2.

=> 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/2.

Ủng hộ mk nha mina^^

cac ban cho mk biet tai sao lai co phan so \(\frac{1}{30};\frac{1}{45}\)vay ???