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\(\frac{2^2}{15}+\frac{2^2}{35}+\frac{2^2}{63}+\frac{2^2}{99}+\frac{2^2}{143}=2\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\right)=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)=2\cdot\left(\frac{1}{3}-\frac{1}{13}\right)=2\cdot\frac{10}{39}=\frac{20}{39}\)
\(=2\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=2\left(1-\frac{1}{13}\right)=2.\frac{12}{13}=\frac{24}{13}\)
\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)
\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)+\left(1-\frac{1}{143}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)
\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=6-\left(1-\frac{1}{13}\right)\)
\(=6-1+\frac{1}{13}\)
\(=5+\frac{1}{13}\)
\(=\frac{66}{13}\)
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
\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{19\times21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{7}{21}-\frac{1}{21}\)
\(=\frac{6}{21}\)
Rút gọn kết quả là \(\frac{2}{7}\), k mk nha mk trả lời đầu tiên đó
\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(A=\frac{2}{5.7}+\frac{4}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(A=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(A=\frac{2}{5.7}+\frac{4}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(A=1.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)\)
A = 1/5 - 1/25
A = 4/25
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(A=2.\frac{3}{16}\)
\(A=\frac{3}{8}\)
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(B=\frac{1}{3}-\frac{1}{21}\)
\(B=\frac{2}{7}\)
Ta có \(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{2}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{2}{7}\)
Vậy \(B=\frac{2}{7}\)
Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
B=2/15+2/35+2/63+2/99
B=2(1/3.5+1/5.7+1/7.9+1/9.11) khoảng cách từ 3-5;5-5;7-9;9-11 là 2 nen
B=2/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11) gop -1/5+1/5;-1/7+1/7;-1/9+1/9=0
B=1(1/3-1/11)=8/33
=2/3*5+2/5*7+2/7*9+2/9*11
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11
=1/3-1/11
=8/33
Ta có:
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)
các bạn.
\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{575}\)
\(\Rightarrow A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{23.25}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{25}\)
\(\Rightarrow A=\frac{25}{75}-\frac{3}{75}\)
\(\Rightarrow A=\frac{22}{75}\)
\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{575}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{23.25}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{3}-\frac{1}{25}\)
\(=\frac{22}{75}\)
Study well ! >_<