Ta có: \(C=\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{51.52}\)C bé hơn\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{50.52}=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{50}-\frac{1}{52}\right)\)

C bé hơn \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{52}\right)\)bé hơn\(\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\)(đpcm)

xin lỗi nha mk ko biết viết kí hiệu bé hơn

mik cảm ơn nha

b, \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

.............................................

Cộng với vế 99 của BĐT trên, ta được:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99.\frac{1}{10}=\frac{99}{10}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}=\frac{1}{10}=\frac{100}{10}=10\)

Wrecking Ball đã làm đúng

to ra kết quả giống cậu : Wrecking Ball

là đáp án đúng

tk nha ( chúc bn học gioi )

k cho đi chế ơi

bằng -9/10

\(=\frac{-9}{10}\)

C = -1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 -1/6 - 1/2  

C = -1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12+1/6 +1/2 ) ( áp dụng quy tắc dấu ngoặc ở lớp 6 ) 

C = -1/90 - [ 1/ ( 9 . 8 ) + 1/ ( 8 . 7 ) + 1/ ( 6 . 7 ) + ...+ 1/(3 .4 ) + 1/ ( 2 . 3 ) + 1/ ( 1. 2 ) ] 

C =  -1/90 - (     1 - 1/2 + 1/2 -1 /3 + .. + 1/7 - 1/8 + 1/8 - 1/9 ) 

C =  -1/90 - ( 1 - 1/9 ) 

C =  -1/90 - 8/9 

C = -1/90 - 80/90

C =   -81/90

C =   -9/10 

Chúc bạn học giỏi 

Tính (9,1 + 10,2 + 11,3 + ...+ 17,9) + 19 = (9,1 + 17,9). 9 : 2 + 19 = 121,5 + 19 = 140,5

Tính Mẫu =  \(\frac{40}{42}.\frac{54}{56}.\frac{70}{72}.....\frac{2650}{2652}=\frac{\left(5.8\right).\left(6.9\right).\left(7.10\right)....\left(50.53\right)}{\left(6.7\right).\left(7.8\right).\left(8.9\right)....\left(51.52\right)}=\frac{\left(5.6.7...50\right).\left(8.9.10....53\right)}{\left(6.7.8...51\right).\left(7.8.9...52\right)}=\frac{5.53}{51.7}=\frac{265}{357}\)

Vậy \(P=\frac{140,5}{\frac{265}{357}}.\frac{265}{357}=140,5\)

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}\)

\(=\frac{8-3}{24}=\frac{5}{24}\)

1/12+1/20+1/30+1/42+1/56=5/24

t.i.c.k nha

Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}\)

\(=\frac{5}{14}\)

Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{1}{6}-\frac{1}{22}\)

\(=\frac{4}{33}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(B=\frac{1}{2}-\frac{1}{7}\)

\(B=\frac{5}{14}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)