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B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101
B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101
B= 1/3 - 1/101
B=98/303
( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )
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Ta có;\(\frac{4}{1\times3}+\frac{4}{3\times5}+\frac{4}{5\times7}+....+\frac{4}{19\times21}\)
\(=2\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+....+\frac{2}{19\times21}\right)\)
\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=2\times\left(1-\frac{1}{21}\right)=2\times\frac{20}{21}=\frac{40}{21}\)
4/1 x 3 + 4/ 3 x 5 + 4/ 5 x 7 + ....+ 4/ 17 x 19 + 4/ 19 x 21
= 2 x ( 2/ 1 x 3 + 2/ 3 x 5 + 2/ 5 x 7 + ...+ 2/ 17 x 19 + 2/ 19 x 21 )
= 2 x ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/17 - 1/19 + 1/19 - 1/21 )
= 2 x ( 1 - 1/21 )
= 2 x 20/21
= 40/21
Chúc bạn học giỏi !!!
Đặt A = 1/3.5 + 1/5.7 + 1/7.9 + ..... + 1/99.101
=> 2A = 2/3.5 + 2/5.7 + 2/7.9 + ..... + 2/99.101
=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101
=> 2A = 1/3 - 1/101
=> 2A = 88/303
=> A = 44/303
Đặt A = 1/3×5 + 1/5×7 + 1/7×9 + ... + 1/97×99
2A = 2/3×5 + 2/5×7 + 2/7×9 + ... + 2/97×99
2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
2A = 1/3 - 1/99
2A = 32/99
A = 32/99 : 2
A = 32/99 × 1/2 = 16/99
A = \(\dfrac{4}{1\times3}\) - \(\dfrac{8}{3\times5}\) + \(\dfrac{12}{5\times7}\) - \(\dfrac{16}{7\times9}\) + \(\dfrac{20}{9\times11}\) - \(\dfrac{24}{11\times13}\)
A = ( \(\dfrac{1}{1}+\dfrac{1}{3}\)) - ( \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\)+ \(\dfrac{1}{7}\)) - ( \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)) +( \(\dfrac{1}{9}\)+ \(\dfrac{1}{11}\)) - (\(\dfrac{1}{11}\)+\(\dfrac{1}{13}\))
A = \(\dfrac{1}{1}+\dfrac{1}{3}\) - \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{12}{13}\)
cho tao chết đi nhá
Đặt \(A=\frac{4}{3\times5}+\frac{4}{5\times7}+\frac{4}{7\times9}+...+\frac{4}{99\times101}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)
\(\Leftrightarrow\frac{A}{2}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Leftrightarrow\frac{A}{2}=\frac{1}{3}-\frac{1}{101}\)
\(\Leftrightarrow\frac{A}{2}=\frac{98}{303}\)
\(\Leftrightarrow A=\frac{98}{303}\times2\)
\(\Leftrightarrow A=\frac{196}{303}\)