P= \(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.17}\)+...+\(\dfrac{3}{96.101}\)

\(\dfrac{5}{3}\).P= \(\dfrac{5}{3}\).(\(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+...+\(\dfrac{3}{96.101}\))

\(\dfrac{5}{3}\).P= \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+...+\(\dfrac{5}{96.101}\)

\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{16}\)+...+\(\dfrac{1}{96}\)-\(\dfrac{1}{101}\)

\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{101}\)= \(\dfrac{101}{101}\)-\(\dfrac{1}{101}\)=\(\dfrac{100}{101}\)

P= \(\dfrac{100}{101}\):\(\dfrac{5}{3}\)= \(\dfrac{100}{101}\).\(\dfrac{3}{5}\)=\(\dfrac{100.3}{101.5}\)=\(\dfrac{20.3}{101.1}\)=\(\dfrac{60}{101}\)

Vậy P= \(\dfrac{60}{101}\)

sửa đề : 

\(E=\frac{3}{1\cdot6}+\frac{3}{6\cdot11}+\frac{3}{11\cdot16}+...+\frac{3}{96\cdot101}\)

\(E=3\left(\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{96\cdot101}\right)\)

\(E=3\left[\frac{1}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{96\cdot101}\right)\right]\)

\(E=3\left[\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\right]\)

\(E=3\left[\frac{1}{5}\left(1-\frac{1}{101}\right)\right]\)

\(E=3\left[\frac{1}{5}\cdot\frac{100}{101}\right]\)

\(E=3\cdot\frac{20}{101}\)

\(E=\frac{60}{101}\)

\(A=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(A=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)

\(A=\frac{3}{5}.\frac{100}{101}\)

\(A=\frac{60}{101}\)

A = 3 - 3/6 + 3/6 - 3/11 + ... + 3/96 - 3/101

A= 3 - 3/101 

A= 300/101

A=\(3.\left(\frac{1}{1.6}\right)+\left(\frac{1}{6.11}\right)+...+\left(\frac{1}{96.101}\right)\)

5A=\(3.\left(\frac{5}{1.6}\right)+\left(\frac{5}{6.11}\right)+...+\left(\frac{5}{96.101}\right)\)

5A=3. \(\left(1-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+...+\left(\frac{1}{96}-\frac{1}{101}\right)\)

5A=3.(1-\(\frac{1}{101}\))

5A=3.\(\frac{100}{101}\)

5A=\(\frac{300}{101}\) suy ra A= \(\frac{300}{101}:5\)=\(\frac{60}{101}\)

\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(\Leftrightarrow B=\frac{3}{5}.\frac{100}{101}\)

\(\Leftrightarrow B=\frac{60}{101}\)

Cảm ơn nhé

Thank you

\(.S=3.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\right)\)

\(\Rightarrow S=3.\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(\Rightarrow S=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(\Rightarrow S=\frac{3}{5}.\left(\frac{100}{101}\right)\)

\(S=\frac{60}{101}\)

\(\frac{100}{101}\)nha

bạn tự tính

tíc mình nha

S=3/1.6+3/6.11+3/11.16+...+3/96.101

=>S=1/1.6+1/6.11+1/11.16+...+1/96.101

S=1-1/6+1/6-1/11+1/11-1/16+...+1/96-1/101

S=1-1/101

S=100/101

\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)

\(=3.\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)\)

\(=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)

\(=\frac{3}{5}.\frac{100}{101}\)

\(=\frac{60}{101}\)

đm dễ thế này thi tự làm đi hỏi cc

Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)

\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)

\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{101}{101}-\dfrac{1}{101}\)

\(=\dfrac{101-1}{101}\)

\(=\dfrac{100}{101}\)