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\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{14}{15}\)
\(=\frac{7}{15}\)
Sửa đề chút nhé:
\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).0\)
\(=0\)
Ý b tham khảo bài bạn nguyen thi thuy linh nhé
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)
\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)+\left(1-\frac{1}{143}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)
\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=6-\left(1-\frac{1}{13}\right)\)
\(=6-1+\frac{1}{13}\)
\(=5+\frac{1}{13}\)
\(=\frac{66}{13}\)
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
Đặt \(A=\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
\(A=\frac{2^{19}\cdot\left(3^3\right)^3+15\cdot\left(2^2\right)^9\cdot\left(3^2\right)^4}{6^9\cdot2^9\cdot2+12^{10}}\)
\(A=\frac{2^{19}\cdot3^9+15\cdot2^{18}\cdot3^8}{12^9\cdot2+12^9\cdot12}=\frac{\left(2^{18}\cdot3^8\right)\cdot6+\left(2^{18}\cdot3^8\right)\cdot15}{12^9\cdot\left(2+12\right)}\)
\(A=\frac{\left(2^{18}\cdot3^8\right)\cdot\left(6+15\right)}{12^9\cdot14}=\frac{2^{18}\cdot3^8\cdot21}{12^9\cdot14}=\frac{2^{18}\cdot3^8\cdot7\cdot3}{2^{18}\cdot3^9\cdot7\cdot2}=\frac{3^8\cdot3}{3^8\cdot3\cdot2}\)
\(A=\frac{1}{2}\)
Đặt \(B=\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}+\frac{4}{195}=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+\frac{4}{11\cdot13}+\frac{4}{13\cdot15}\)
\(B=\frac{1}{2}\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{13}-\frac{4}{15}\right)\)
\(B=\frac{1}{2}\left(\frac{4}{5}-\frac{4}{15}\right)\)mà \(\frac{4}{5}-\frac{4}{15}< 1\Leftrightarrow\frac{1}{2}\left(\frac{4}{5}-\frac{4}{15}\right)< \frac{1}{2}\Leftrightarrow B< A\)
