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A=1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999
A= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ...+ 1/99x101
Ax2= 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101
Ax2= 5-3/3x5 + 7-5/5x7 + 9-7/7x9 + 11-9/9x11 + ... + 101-99/99x101
Ax2=5/3x5 - 3/3x5 + 7/5x7 - 5/5x7 + 9/7x9 -7/7x9 + ... + 101/99x101 -99/99x101
Ax2=1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/99 - 1/101
Ax2= 1/3 - 1/101
Ax2 = 98/303
A= 98/303 : 2
A=49/303
mình không biết nữa bằng bao nhiêu ấy nhỉ .......? .......? Sory ^.^
1/3 + 13/15 + 33/35 + 61/63 + 97/99
= 45/11 ( mình không tiện giải, để khi khác giải sau)
Chúc bạn may mắn!
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\times\frac{98}{303}\)
\(A=\frac{49}{303}\)
A= \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ ... + \(\frac{1}{9999}\)
A= \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\) + ... + \(\frac{1}{99.101}\)
2. A= \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) + ... + \(\frac{2}{99.101}\)
2.A = \(\frac{1}{3}\) - \(\frac{1}{5}\)+ \(\frac{1}{5}\)-\(\frac{1}{7}\) + ... + \(\frac{1}{99}\) - \(\frac{1}{101}\)
2.A= \(\frac{1}{3}\) - \(\frac{1}{101}\)
2.A= \(\frac{101}{303}\) - \(\frac{3}{303}\)
2.A= \(\frac{98}{303}\)
A = \(\frac{98}{303}\) : 2
A = \(\frac{49}{303}\)
Vay A=\(\frac{49}{303}\)
\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\frac{10}{11}.y=\frac{4}{3}\)
\(\Rightarrow y=\frac{22}{15}\)
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{1}{3}+\dfrac{1}{25}\)
\(=\dfrac{28}{75}\)
= 1/(3x5) + 1/(5x7) + 1/(7x9) +.....+ 1/(99x101) =( 1/3 -1/5 + 1/5 -1/7 +1/7 - 1/9 +....+ 1/99 -1/101 ) :2 = (1/3 -1/101) : 2 = 98/303 : 2 = 49/303
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}\)
\(=\frac{49}{303}\)
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