a,1/1-1/4+1/4-1/7+...+1/2008-1/2011

=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)

=1-1/2011+0+...+0

=1-1/2011

=2010/2011

Q=5(5/1x6+5/6x11+5/11x16+....+5/26x31)

Q=5(1/1-1/6+1/6-1/11+1/11-1/16+....+1/26-1/31)

Q=5(1/1-1/31)

Q=5x30/31

Q=150/31

\(Q=\frac{25}{1.6}+\frac{25}{6.11}+\frac{25}{11.16}+......+\frac{25}{26.31}.\)

\(Q=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{26}-\frac{1}{31}\right)\)

\(Q=5\left(1-\frac{1}{31}\right)\)

CÒN ĐÔU PN TỰ LÀM NHA

\(Q=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)

\(Q=5\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{26.31}\right)\)

\(Q=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(Q=5\left(1-\frac{1}{31}\right)\)

\(Q=5.\frac{31}{32}\)

\(Q=\frac{155}{32}\)

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

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

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

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

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

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

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

C=5(5/1.6+5/6.11+...+5/26.31)

C=5(1-1/6+1/6-1/11+...+1/26-1/31)

C=5(1-1/31)

C=5.30/31

C=150/31

CHÚC BẠN HỌC TỐT!!!

\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{1001.1006}\)

\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{1001.1006}\right)\)

\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{1001}-\frac{1}{1006}\right)\)

\(=5\left(1-\frac{1}{1006}\right)\)

\(=5.\frac{1005}{1006}\)

\(=\frac{5025}{1006}\)

                                 \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{1001.1006}\)

                      \(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{1001.1006}\right)\)

                     \(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{1001}-\frac{1}{1006}\right)\)

                  \(=5.\left(1-\frac{1}{1006}\right)\)

                  \(=\frac{5.1005}{1006}=\frac{5025}{1006}\)

                 Ủng hộ mk nha !!! ^_^

Ta có : \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+.....+\frac{5^2}{26.31}\)

\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{26.31}\right)\)

\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{31}\right)\)

\(=5\left(1-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)

E=\(\frac{10}{1\cdot6}\) +\(\frac{10}{6\cdot11}\) +\(\frac{10}{11\cdot16}\) +\(\frac{10}{16\cdot21}\) +\(\frac{10}{21\cdot26}\) +\(\frac{10}{26\cdot31}\)                                                                                      =  5*(1-\(\frac{1}{31}\) )                                                                                                                                                                                 =5*\(\frac{30}{31}\)        =\(\frac{150}{31}\) 

=> 1 - 1/6 + 1/6 - 1/11 +.......+1/5x+1 - 1/5x+6=2010/2011

=> 1 - 1/5x+6 = 2010/2011

=> 1/5x+6 = 1/2011

=> 5x + 6 = 2011

=> 5x = 2005

=> x = 401(tm)