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

\(A=\frac{10^2}{1\cdot6}+\frac{10^2}{6\cdot11}+...+\frac{10^2}{61\cdot66}=\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{61\cdot66}\right)\cdot20\)

\(=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{61}-\frac{1}{66}\right)\cdot20\)

\(=\left[\left(1-\frac{1}{66}\right)+\left(\frac{1}{6}-\frac{1}{6}\right)+...+\left(\frac{1}{61}-\frac{1}{61}\right)\right]\cdot20\)

\(=\left[\left(\frac{66}{66}-\frac{1}{66}\right)+0+...+0\right]\cdot20=\frac{65}{66}\cdot20=\frac{65\cdot20}{66}=\frac{65\cdot10}{33}=\frac{650}{33}\)

\(A=\frac{10^2}{1.6}+\frac{10^2}{6.11}+...+\frac{10^2}{61.66}\)

\(=10^2.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{61.66}\right)\)

\(=10^2.5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{61}-\frac{1}{66}\right)\)

\(=500.\left(1-\frac{1}{66}\right)\)

\(=500.\frac{65}{66}\)

\(=\frac{16250}{33}\)

Tính S:

S=5.(\(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+...+\(\dfrac{5}{101.106}\))

S=5.(1-\(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\))

S=5.(1-\(\dfrac{1}{106}\))

S=5.\(\dfrac{105}{106}\)

S=\(\dfrac{525}{106}\)

525/106

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

\(=\dfrac{10}{5}.\left(\dfrac{1}{1.6}+\dfrac{1}{6.11}+...+\dfrac{1}{101.106}\right)\)

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

\(=2.\left(\dfrac{1}{1}-\dfrac{1}{106}\right)\)

\(=2.\dfrac{105}{106}\)

= \(\dfrac{2.105}{106}\)\(=\dfrac{210}{106}=\dfrac{105}{53}\)

 = 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

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

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

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

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

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

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

2 phần dưới không liên quan gì đến tính chất trên 
a) \(A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{20-17}{17.20}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
b) \(B=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{106-101}{101.106}\right)\)
\(B=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{106}\right)\)
\(B=5.\left(1-\frac{1}{106}\right)=\frac{525}{106}\)

Có liên quan đó bạn!

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 

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

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

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

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

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

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