ta co:

2A=2(2 mu 60 +1 /2 mu 61 +1)

2A=2 mu 61 +2 / 2 mu 61 +1

2A=2 mu 61 +1+1/2 mu 61 +1

2A=1+1/2 mu 61 +1

ta co:

2B=2(2 mu 61 +1/2 mu 62 +1)

2B=2 mu 62 +2/2 mu 62+1

2B=2 mu 62 +1+1/2 mu 62 +1

2B=1+1/2 mu 62 +1

mà 1+1/2 mu 61+1>1+1/2 mu 62 +1 nen 2A >2B

vậy A>B

nhớ k đúng cho mk nha

Ta có:

2.A=2 mủ 61 +2/2 mủ 61 +1=1+(2/2 mủ 61 +1)

2.B=2 mủ 62 + 2 /2 mủ 62 +1=1+(2/2 mủ 62 + 1)

vì ... nên 2.A >2.B.Vậy A>B

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{7^2}\right)\)

\(< \frac{1}{2^2}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{1}{2^2}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2^2}\cdot\frac{6}{7}\)

\(=\frac{3}{14}\)

\(< \frac{1}{2}\)

Bạn tính sao cho bằng nhau rồi so sánh là xong

mk cá rằng các số trong ô vàng nhỏ hơn 1 phần 2 là chắc vì kết quả dãy dài đó là: 0.477727816. kết bạn với mk nha.

1/ 2 lớn hơn

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 / 30) * 5 = 1 / 3 + 1 / 6 = 
(1 / 6) x (2 + 1) = (1 / 6) x  3 = 1/2

cái này mà là PS

Ta có :

\(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{4}\)

\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{1}{20}\)

\(\Rightarrow D=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy \(D< \frac{1}{2}\)

\(D=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)

Nhận xét: \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)

\(\Rightarrow D< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy D < 1/2

a: so sánh với 1

64/85 < 73/81

b: so sánh với 1 

n + 1/n+2 > n/ n+3

c: so sánh với 1

64/65 > 60/61

d: so sánh với 1

99/97 < 88/86

\(A=\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)

* Cách 1 : 

Ta có : 

\(5A=\frac{5^{61}+5}{5^{61}+1}=\frac{5^{61}+1+4}{5^{61}+1}=\frac{5^{61}+1}{5^{61}+1}+\frac{4}{5^{61}+1}=1+\frac{4}{5^{61}+1}\)

\(5B=\frac{5^{62}+5}{5^{62}+1}=\frac{5^{62}+1+4}{5^{62}+1}=\frac{5^{62}+1}{5^{62}+1}+\frac{4}{5^{62}+1}=1+\frac{4}{5^{62}+1}\)

Vì \(\frac{4}{5^{61}+1}>\frac{4}{5^{62}+1}\) nên \(1+\frac{4}{5^{61}+1}>1+\frac{4}{5^{62}+1}\) 

\(\Rightarrow\)\(5A>5B\) hay \(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~