Sửa đề: \(98+99+\dfrac{142}{144}\) \(\rightarrow\dfrac{98}{99}+\dfrac{143}{144}\)  

Giải:

\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+\dfrac{62}{63}+\dfrac{98}{99}+\dfrac{143}{144}+\dfrac{194}{195}\) 

\(A=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{195}\right)\) 

\(A=7-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{195}\right)\) 

\(A=7-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\dfrac{14}{15}\right]\) 

\(A=7-\dfrac{7}{15}\) 

\(A=\dfrac{98}{15}\) 

Chúc bạn học tốt!

sửa đề: \(B=5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}\)

Ta có: \(A=\frac23+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}+\frac{194}{195}\)

\(=1-\frac13+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}+1-\frac{1}{143}+1-\frac{1}{195}\)

\(=7-\left(\frac13+\frac{1}{15}+\cdots+\frac{1}{195}\right)\)

\(=7-\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{13\cdot15}\right)\)

\(=7-\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{13}-\frac{1}{15}\right)=7-\frac12\left(1-\frac{1}{15}\right)\)

\(=7-\frac12\cdot\frac{14}{15}=7-\frac{7}{15}=\frac{98}{15}\) >6

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{7^2}<\frac{1}{6\cdot7}=\frac16-\frac17\)

Do đó; \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<1-\frac12+\frac12-\frac13+\cdots+\frac16-\frac17=1-\frac17<1\)

=>\(5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<5+1=6\)

=>B<6

mà A>6

nên B<A

\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)

\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)+\left(1-\frac{1}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)

\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

Mk sửa lại 1 tí nha dòng thứ 5 :

\(A=6-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}\left(1-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}.\frac{12}{13}\)

\(=6-\frac{6}{13}=\frac{72}{13}\)

Mong bn bỏ qua nha

\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)

\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)

\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)

\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)

\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)

\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)

\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)

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Đâu rồi

cậu k đi chơi trung thu à

Ko mìk ko đi chơi bạn giải dc câu nào ko giải giúp mìk vs

Bài 1:

1: \(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{11\cdot13}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{11\cdot13}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac12\left(\frac13-\frac{1}{13}\right)=\frac12\cdot\frac{10}{39}=\frac{5}{39}\)

2: \(C=\frac12+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(=\frac24+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}+\frac{2}{304}\)

\(=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}+\frac{2}{16\cdot19}\)

\(=\frac23\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\right)\)

\(=\frac23\left(1-\frac14+\frac14-\frac17+\frac17-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(=\frac23\left(1-\frac{1}{19}\right)=\frac23\cdot\frac{18}{19}=\frac{2\cdot6}{19}=\frac{12}{19}\)

Bài 2:

\(\frac{1}{101}>\frac{1}{300};\frac{1}{102}>\frac{1}{300};\ldots;\frac{1}{299}>\frac{1}{300};\frac{1}{300}=\frac{1}{300}\)

Do đó: \(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+\cdots+\frac{1}{300}=\frac{200}{300}=\frac23\) (ĐPCM)

\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{5}{39}\)