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=(1+2+3+4+5+6+...)x(9x11+1-100)
=(1+2+3+4+5+6+...)x99+1-100
=(1+2+3+4+5+6+...)x100-100
=(1+2+3+4+5+6+...)x0
=0
(1+2+3+4+5+6...)x(9x11-100+1)
=(1+2+3+4+5+6+...)x99+1-100
=(1+2+3+4+5+6+...)x100-100
=(1+2+3+4+5+6+...)x0
=0
Vậy kết quả của biểu thức trên bằng 0
\(\frac{4}{9\times11}+\frac{4}{11\times13}+\frac{4}{13\times15}+...+\frac{4}{95\times97}+\frac{4}{97\times99}\)
\(=2\times\left(\frac{2}{9\times11}+\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{95\times97}+\frac{2}{97\times99}\right)\)
\(=2\times\left(\frac{11-9}{9\times11}+\frac{13-11}{11\times13}+\frac{15-13}{13\times15}+...+\frac{97-95}{95\times97}+\frac{99-97}{97\times99}\right)\)
\(=2\times\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\times\left(\frac{1}{9}-\frac{1}{99}\right)=\frac{20}{99}\)
4/9x11 + 4/11x13 + 4/13X15 + .............+ 4/95X97 + 4/97X99
=2 x (2/9x11 + 2/11x 13 + .........+2/95x97 + 2/97x99)
=2 x ( 1/9 - 1/11 + 1/11- 1/13 +...... + 1/97 - 1/99)
=2 x (1/9 - 1/99)
=2 x10/99
=20/99
Học tốt!
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}\right)\)
\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{11}{11}-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)
Ta có:
A = \(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
= \(\frac{1}{1}-\frac{1}{11}\)
=\(\frac{10}{11}\)
9x10=90
so can tim la 90
9x10=90