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A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{129.15}\)
A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{129.15}\)
A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{129}-\frac{1}{15}\)
A = \(\frac{1}{3}-\frac{1}{15}\)= \(\frac{5}{15}-\frac{1}{15}\)
A = \(\frac{4}{15}\)
CẢM ƠN LỜI GỢI Ý
Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57
=1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)
=1/2(1/3-1/57)
=1/2(19/57-1/57)
=1/2.18/57
=3/19
Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19
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Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}\)
\(=\frac{49}{303}\)
Dấu chấm(.) ở cấp hai là dấu nhân (x)
a)Ta có:
A= 1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A= 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39
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
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)
\(A=\frac{1}{3}-\frac{1}{101}\)
\(A=\frac{98}{303}\)
A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
= \(\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{99.101}\right):2\)\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
= \(\left(\frac{1}{3}-\frac{1}{101}\right):2=\frac{101-3}{303}:2=\frac{98}{303}:2=\frac{49}{303}\)
Dấu chấm trong bài là dấu nhân nha !
A = 1/15 + 1/35 + 1/63 + 1/99 +........1/9999
A = 1/3x5 + 1/5x7 + 1/7x9 + .......+ 1/99x101
A = 1/2 x (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +..... + 1/99 - 1/101)
A = 1/2 X ( 1/3 - 1/99)
A = 1/2 x ( 1/3 - 1/101)
A = 1/2 ( 98/303)
A = 49/303
1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999 =
= 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
ĐS: 49/303
Tick nha
1/3x5 +1/5x7+1/7x9 +1/9x11+...+1/99x101
1/3-1/5+1/5-1/7+...+1/99-1/101
1/3-1/101
98/303