Lời giải:

$12A=1.5.12+5.9.(13-1)+9.13(17-5)+13.17(21-9)+....+77.81(85-73)+81.85(89-77)$

$=60+(5.9.13+9.13.17+13.17.21+...+77.81.85+81.85.89)-(1.5.9+5.9.13+9.13.17+...+73.77.81+77.81.85)$

$=60+81.85.89 - 1.5.9=612780$

A = 1.5 + 5.9 + 9.13 + ... + 81.85

A = \(\dfrac{12}{12}\)(1.5 + 5.9 + 9.13 + 81.85)

A = \(\dfrac{1}{12}\).(1.5.12 + 5.9.12.+ 9.13.12 + ...+ 81.85.12]

A = \(\dfrac{1}{12}\).[1.5.(9 + 3) + 5.9.(13 - 1) + 9.13.(17 - 5) +...+ 81.85.(89 - 77)]

^{A = \(\dfrac{1}{12}\).[1.5.9 + 1.3.5 + 5.9.13 - 5.9.1 + 9.13.17 - 9.13.5 + ...+ 81.85.89 - 81.85.77]}

A = \(\dfrac{1}{12}\).[1.3.5 + 81.85.89]

A = \(\dfrac{1}{12}\).[15 + 612765]

A = \(\dfrac{1}{12}\).612780

A = 51065

5644/45

\(A=\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+....+\frac{2}{81.85}\)

\(\Rightarrow2A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{81.85}\)

\(\Rightarrow2A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{85}\)

\(\Rightarrow2A=1-\frac{1}{85}\)

\(\Rightarrow A=\frac{84}{85}:2=\frac{42}{85}\)

tính A còn lại tự tính nha

a) A = 2/1x5 + 2/5x9 + 2/9x13 +....+2/81x85

\(\frac{2}{1x5}+\frac{2}{5x9}+\frac{2}{9x13}+...+\frac{2}{81x85}\)

\(A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{85}\)

\(A=1-\frac{1}{85}\)

\(\Rightarrow A=\frac{84}{85}\)

k nha

A = \(\frac{2}{1\cdot5}\) + \(\frac{2}{5\cdot9}\) + \(\frac{2}{9\cdot13}\) + ... + \(\frac{2}{81\cdot85}\)

   = \(\frac{1}{2}\) *  ( \(\frac{1}{1\cdot5}\) + \(\frac{1}{5\cdot9}\) + \(\frac{1}{9\cdot13}\) + ... + \(\frac{1}{81\cdot85}\) )

   = \(\frac{1}{2}\) * ( 1 - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{9}\) + \(\frac{1}{9}\) - \(\frac{1}{13}\) + ... + \(\frac{1}{81}\) - \(\frac{1}{85}\) )

   = \(\frac{1}{2}\) * ( 1 - \(\frac{1}{85}\) )

   = \(\frac{1}{2}\) * \(\frac{84}{85}\)

   = \(\frac{42}{85}\)

A = 4/1 x 5 + 4/5 x 9 + 4/9 x 13 + .... + 4/91 x 95 + 4/95 x 99

A = 1 - 1/5 + 1/5 -1/9 + 1/9 - 1/13 + .... + 1/91 - 1/95 + 1/95 - 1/99

A = 1 - 1/99

A = 98/99

B = 1/6 + 1/12 + 1/20 + ... + 1/132

B = 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + ... + 1/11 x 12

B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/11 - 1/12 

B = 1/2 - 1/12 

B = 5/12

a) \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{47.49}=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{47.49}\right)=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)

\(\Rightarrow\frac{24}{49}=\frac{24}{x+4}\)

\(\Rightarrow x+4=49\Rightarrow x=45\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{49}\right)=\frac{24}{x+4}\Leftrightarrow A=\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{24}{49}=\frac{24}{x+4}\Leftrightarrow x=49-4=45\)

Bài b) hình như sai đề thì phải đó bạn.

a)ta co \(\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{47.49}=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{24}{49}=\frac{24}{x+4}\)\(\Rightarrow49=x=4\Rightarrow x=45\)

b)ta co \(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{97.101}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}=\frac{2x+4}{101}\Rightarrow100=2x+4\Rightarrow x=48\)