S = 2(5/1.6 + 5/6.11 +.......+ 5/101.106)

S = 2( 1 - 1/6 + 1/6 - 1/11 +.....+ 1/101 - 1/106)

S = 2( 1 - 1/106)

S = 2 . 105/106

S = 105/53

k mk đi,mk mới bị trừ điểm!

1/2.S=5/(1.6)+5/(6.11)+...+5/(101.106)

1/2.S=1/1-1/6+1/6-1/11+...+1/101-1/106

1/2.S=1/1-1/106

1/2.S=105/106

S=105/53

 = 2.(5/1.6+5/6.11+.....+5/56.61)

 = 2.(1-1/6+1/6-1/11+.....+1/56-1/61)

 = 2.(1-1/61)

 = 2.60/61 = 120/61

Tk mk nha

\(\frac{10}{1.6}+\frac{10}{6.11}+...+\frac{10}{56.61}\)

\(=10.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{56.61}\right)\)

\(=10.\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(=2\left(1-\frac{1}{61}\right)\)

\(=2.\frac{60}{61}\)

\(=\frac{120}{61}\)

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

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

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

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

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

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

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

\(S=2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(S=2\left(1-\frac{1}{106}\right)\)

\(S=\frac{210}{106}=\frac{105}{53}\)

\(S=2.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(S=2.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{106}\right)\)

\(S=2.\left[1+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{11}\frac{1}{11}\right)+...+\left(\frac{-1}{101}+\frac{1}{101}\right)-\frac{1}{106}\right]\)

\(S=2.\left[1+0+0+...+0-\frac{1}{106}\right]\)

\(S=2.\left[1-\frac{1}{106}\right]\)

\(S=2.\frac{105}{106}\)

\(S=\frac{10}{1.6}+\frac{10}{6.11}+...+\frac{10}{101.106}\)

=\(2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(=2\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}+\frac{1}{106}\right)\)

\(=2\left(1-\frac{1}{106}\right)=2.\frac{105}{106}=\frac{105}{53}\)

Ta có :

\(\frac{5}{1.6}+\frac{5}{6.11}+................+\frac{5}{\left(5.x+1\right).\left(5.x+6\right)}=\)\(\frac{50}{41}\)

=> \(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...............+\frac{1}{5.x+1}-\frac{1}{5.x+6}\) = \(\frac{50}{41}\)

=> \(1-\frac{1}{5.x+6}=\frac{50}{41}\)

=> \(\frac{1}{5.x+6}=\frac{-9}{41}\)................ mình ko tìm ra vì p/s kia ko có tử là 1

bạn xem lại đề bài giúp mình nha 

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

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

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

\(a\)) Mình giải theo cách khác:

Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)

Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

A = \(\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{9}.\frac{1}{10}\)

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

A = \(1-\frac{1}{10}\)

A = \(\frac{9}{10}\)

1/2=1-1/2 ; 1/2.1/3=1/2-1/3 ; 1/3.1/4=1/3-1/4...v...v

Vậy A bằng: 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.............+1/8-1/9+1/9-1/10

                =1-1/10=9/10