\(A=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}\right):5\)

\(A=\left(1-\frac{1}{501}\right):5\)

\(A=\frac{500}{501}:5=\frac{100}{501}\)

Ta có : \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)

    \(\Rightarrow\)  \(A=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{496}-\frac{1}{501}\right) \)

     \(\Rightarrow\)  \(A=\frac{1}{5}\left(1-\frac{1}{501}\right)\)

      \(\Rightarrow\)  \(A=\frac{1}{5}.\frac{501-1}{501}=\frac{1}{5}.\frac{500}{501}\)

       \(\Rightarrow\)  \(A=\frac{1.500}{5.501}=\frac{20}{1.501}=\frac{20}{501}\)     

                                               Vậy   \(A=\frac{20}{501}\)

mk nhầm 1 chút : \(A=\frac{1.100}{5.101}=\frac{100}{1.101}=\frac{100}{101}\)

                         Vậy   \(A=\frac{100}{101}\) chứ ko phải bằng  \(\frac{20}{101}\)  đâu nhé mong bn thông cảm!!!!

Nhân A với 5 ta được

A .5=5.(1/1.6 +1/6.11+1/11.16+....+1/496.561)

A .5=5/1.6+5/6.11+5/11.16+...+5/496.561

A .5=1-1/6+1/6-1/11+1/11-1/16+...+1/496-1/561

A .5=1-1/561

A .5=560/561

A    =560/561 : 5 =112/561

\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.561}\)

\(=\left(\frac{1}{1}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{16}\right)+...+\left(\frac{1}{496}-\frac{1}{561}\right)\)

\(=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{496}-\frac{1}{561}\)

\(=\frac{1}{1}-\frac{1}{561}\)

\(=\frac{561}{561}-\frac{1}{561}\)

\(=\frac{560}{561}\)

\(\Rightarrow\) Vậy, \(A=\frac{560}{561}\)

=1-1/6+1/6-1/11+1/11-1/16+...+1/496-1/561

=1-1/561

=560/561

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

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

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

\(A=\frac{1}{5}\cdot\frac{100}{101}\)

\(A=\frac{20}{101}\)

A = 1/5(1-1/6+1/6-1/11+1/11-1/16+.....+1/96-1/101)

    = 1/5(1-1/101)=20/101

ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101

= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101

=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)

=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)

=1/5 . (1/1-1/101)

=1/5 . 100/101

= 20/101

5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)

  =\(1- {1 \over 101}={100 \over 101}\)

suy ra A =\({20 \over 101}\)

\(A=\frac{1}{5}\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{...5}{96\cdot101}\right)\)

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

\(A=\frac{1}{5}\left(\frac{1}{1}+0+0+0+...+0-\frac{1}{101}\right)\)

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

\(A=\frac{1}{5}\cdot\frac{100}{101}\)

\(A=\frac{20}{101}\)

Ta có:

Số hạng thứ nhất: \(\frac{1}{1.6}=\frac{1}{\left(5.1-4\right).\left(5.1+1\right)}\)

Số hạng thứ 2: \(\frac{1}{6.11}=\frac{1}{\left(5.2-4\right).\left(5.2+1\right)}\)

Số hạng thứ 3: \(\frac{1}{11.16}=\frac{1}{\left(5.3-4\right)+\left(5.3+1\right)}\)

.......

Số hạng thứ n = \(\frac{1}{\left(5.n-4\right)+\left(5.n+1\right)}\)

Vậy số hạng 100 của dãy đó là: \(\frac{1}{\left(5.100-4\right).\left(5.100+1\right)}=\frac{1}{496.501}\)

\(A=\)\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{51.56}\)

\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{51.56}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{51}-\frac{1}{56}\)

\(5A=1-\frac{1}{56}=\frac{55}{56}\)

\(A=\frac{55}{56}\div5=\frac{55}{56}.\frac{1}{5}=\frac{11}{56}\)

\(A=\dfrac{1}{1\cdot6}-\dfrac{1}{6\cdot11}-\dfrac{1}{11\cdot16}-\dfrac{1}{16\cdot21}-...-\dfrac{1}{46\cdot51}\)

\(=\dfrac{1}{6}-\left(\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+\dfrac{1}{16\cdot21}+...+\dfrac{1}{46\cdot51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{5}{6\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{46\cdot51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{46}-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{51}\right)\)

\(=\dfrac{1}{6}-\dfrac{1}{5}\cdot\dfrac{5}{34}\)

\(=\dfrac{1}{6}-\dfrac{1}{34}\)

\(=\dfrac{7}{51}\)

Vậy \(A=\dfrac{7}{51}\)

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

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