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ta có A =\(\frac{1}{5\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot15}+...+\frac{1}{605\cdot608}\)
3A =\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{605\cdot608}\)
3A =\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\)
3A=\(\frac{1}{5}-\frac{1}{608}\)
3A=\(\frac{603}{3040}\)A =\(\frac{201}{3040}\)
Đặt A=\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\)
3A=\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\right)\)
3A=\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\right)\)
3A=3.\(\left(\frac{1}{5}-\frac{1}{608}\right)\)
A=\(\frac{201}{3040}\)
Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)
\(A=\frac{1}{5}-\frac{1}{605}\)
\(A=\frac{24}{121}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{602\cdot605}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)
\(=\frac{1}{5}-\frac{1}{605}+0+...+0\)
\(=\frac{24}{121}\)
Đề là cm S>1 nha bạn!
\(S=\frac{9}{2.5}+\frac{9}{5.8}+...+\frac{9}{29.32}\)
\(=3\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(=3.\frac{15}{32}\)
\(=\frac{45}{32}>1\)
\(\Leftrightarrow S>1\)
\(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}\)
Cách 1 : Vì hiệu hai thừa số đều là 3 = 5 - 2 = 8 - 5 = ... = 32 - 29 nên phân tích tử 9 = 3 . 3
Ta có : \(S=3\left[\frac{3}{2\cdot5}+\frac{3}{7\cdot9}+...+\frac{3}{29\cdot32}\right]=3\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{32}\right]\)
\(=3\left[\frac{1}{2}-\frac{1}{32}\right]=3\left[\frac{16}{32}-\frac{1}{32}\right]=3\cdot\frac{15}{32}=\frac{45}{32}\)
Mà \(\frac{45}{32}>1\)=> S không thể bé hơn 1
Cách 2 : Nhận xét : \(\frac{9}{2\cdot5}=\frac{3}{2}-\frac{3}{5};\frac{9}{5\cdot8}=\frac{3}{5}-\frac{3}{8};...\)
Vậy ta có : \(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}=\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+...+\frac{3}{29}-\frac{3}{32}\)
\(=\frac{3}{2}-\frac{3}{32}=\frac{3\cdot16}{32}-\frac{3}{32}=\frac{48}{32}-\frac{3}{32}=\frac{45}{32}\)
Tự so sánh , mà S đâu bé hơn 1 ???
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\)\(\frac{3}{11\cdot14}+...+\)\(\frac{3}{602\cdot605}\)
\(=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{602\cdot605}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}\)\(-\frac{1}{605}\)
\(=\frac{1}{5}-\frac{1}{605}\)
\(=\frac{121}{605}-\frac{1}{605}\)
\(=\frac{120}{605}=\frac{24}{121}\)
Bài này dùng công thức nhé
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{602.605}\)
\(=\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}-\frac{1}{605}\)
\(=\)\(\frac{1}{5}-\frac{1}{605}\)
\(=\)\(\frac{24}{121}\)
Chúc bạn học tốt ~
a)\(\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{x}-\frac{1}{x+3}\right)\)=\(\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
=\(\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}\)=\(\frac{101}{1540}:\frac{1}{3}\)=\(\frac{303}{1540}\)
\(\frac{1}{x+3}\)=\(\frac{1}{5}-\frac{303}{1540}\)=\(\frac{1}{308}\)
\(\Rightarrow\)x+3=308
\(\Rightarrow\)x=308-3=305
b)Mk chưa nghĩ ra
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x+1-6}{6\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow3x=51\)
\(\Rightarrow x=17\)
Vậy x = 17
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{48}{98}\)
\(A=\frac{8}{49}\)
A = \(\frac{1}{3}\).{ \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)}
A = \(\frac{1}{3}\).{\(\frac{1}{2}-\frac{1}{98}\)}
A = \(\frac{1}{3}.\left\{\frac{49}{98}-\frac{1}{98}\right\}\)
A=\(\frac{1}{3}.\frac{24}{49}\)
A = \(\frac{49}{98}\)
