đáp án là 277.1937

277.1937

\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)

\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

4/1x3x5+4/3x5x7+4/5x7x9+4/7x9x11+4/9x11x13

=1/1x3-1/3x5+1/3x5-1/5x7+1/5x7-1/7x9+1/7x9-1/9x11+1/9x11-1/11x13

=1/1x3-1/11x13

=1/3-1/143=140/429

giúp mình với ạ

\(\dfrac{4}{1\times3\times5}+\dfrac{4}{3\times5\times7}+\dfrac{4}{5\times7\times9}+\dfrac{4}{7\times9\times11}\)

=\(\dfrac{5-1}{1\times3\times5}+\dfrac{7-3}{3\times5\times7}+\dfrac{9-5}{5\times7\times9}+\dfrac{11-7}{7\times9\times11}\)

=\(\dfrac{1}{1\times3}-\dfrac{1}{3\times5}+\dfrac{1}{3\times5}-\dfrac{1}{5\times7}+...+\dfrac{1}{9\times11}-\dfrac{1}{11\times13}\)

=\(\dfrac{1}{3}-\dfrac{1}{143}=\dfrac{140}{429}\)

\(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)

= \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\frac{1}{7.9.11}+\frac{1}{9.11.13}\)

= \(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

= \(\frac{1}{1.3}-\frac{1}{11.13}\)

= \(\frac{140}{429}\)

~~~

Không chắc chắn lắm nhé :3

#Sunrise

nhớ cho k nhé !

4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13

1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)

Khó hiểu quá

đề bài

\(A=\frac{2.6.10+6.10.14+...+1994.1998.2002}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{1.2^3.3.5+3.2^3.5.7+....+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)

\(=8\)

Vậy A = 8