\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+........+\frac{1}{11x13}+\frac{1}{13x15}\)
ai đúng sẽ k cho người đó
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\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Bài làm :
Ta có :
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{31}{15}\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
Chúc bn hok giỏi !!!!!!!!! ^_^
a) Ta có: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}=1\text{-}\frac{1}{3}+\frac{1}{3}\text{-}\frac{1}{5}+...+\frac{1}{11}\text{-}\frac{1}{13}=1\text{-}\frac{1}{13}=\frac{12}{13}\)
Thay vào ta có:
\(\frac{12}{13}+x=\frac{24}{13}\Rightarrow x=\frac{24}{13}\text{-}\frac{12}{13}\Rightarrow x=\frac{12}{13}\)
\(\frac{2}{1x3}+\)\(\frac{2}{3x5}+\)\(\frac{2}{5x7}+\)\(\frac{2}{7x9}+\frac{2}{9x11}+\frac{2}{11x13}\)
= \(\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\)\(+\frac{13-11}{11x13}\)
= \(\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+\frac{9}{7x9}-\frac{7}{7x9}+\frac{11}{9x11}\)\(-\frac{9}{9x11}\)\(+\frac{13}{11x13}-\frac{11}{11x13}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\)\(\frac{1}{13}\)
= \(1-\frac{1}{13}=\frac{12}{13}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{57.59}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{59}\)
\(B=1-\frac{1}{59}\)
\(B=\frac{59}{59}-\frac{1}{59}=\frac{58}{59}\)
Vậy B = \(\frac{58}{59}\)
Lưu ý: Dấu "." là dấu nhân
\(B=\frac{2}{1.3}+\frac{1}{3.5}+\frac{2}{5.7}+...+\frac{1}{57.59}\)
\(B=1.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{57}-\frac{5}{59}\right)\)
\(B=1.\left(1-\frac{1}{59}\right)\)
\(B=1.\frac{58}{59}\)
\(B=\frac{58}{59}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
\(\Leftrightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}\right)=2.\frac{8}{17}\)
\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
\(\Rightarrow x+2=17\Rightarrow x=15\)
\(=\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}+...+\frac{15-13}{13x15}=\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{13}-\frac{1}{15}=\frac{1}{5}-\frac{1}{15}=\frac{2}{15}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
mik đã trả lời rồi mà , sao chưa hiện ra ????