c/ 1/1.3+ 1/3.5+1/5.7+1/7.9+ 1/9.11 d/ 2/1.3+2/3.5+ 2/5.7+ 2/7.9+ 2/9.11 e/ (1+1/2).(1+1/3).(1+1/4)…(1+1/100) f / (1-1/2).(1-1/3).(1-1/4)…(1-1/100

c/ 1/1.3+ 1/3.5+1/5.7+1/7.9+ 1/9.11
d/ 2/1.3+2/3.5+ 2/5.7+ 2/7.9+ 2/9.11
e/ (1+1/2).(1+1/3).(1+1/4)…(1+1/100)
f / (1-1/2).(1-1/3).(1-1/4)…(1-1/100)

0 bình luận về “c/ 1/1.3+ 1/3.5+1/5.7+1/7.9+ 1/9.11 d/ 2/1.3+2/3.5+ 2/5.7+ 2/7.9+ 2/9.11 e/ (1+1/2).(1+1/3).(1+1/4)…(1+1/100) f / (1-1/2).(1-1/3).(1-1/4)…(1-1/100”

  1. Mình chỉ giải được câu c và d thôi mong bạn thông cảm.

    c) $\frac{1}{1.3}$  + $\frac{1}{3.5}$  + $\frac{1}{5.7}$ + $\frac{1}{7.9}$  + $\frac{1}{9.11}$ 
    ⇔ $\frac{1}{2}$.($\frac{2}{1.3}$  + $\frac{2}{3.5}$  + $\frac{2}{5.7}$ + $\frac{2}{7.9}$  + $\frac{2}{9.11}$ )
    ⇔ $\frac{1}{2}$.($\frac{1}{1}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – $\frac{1}{5}$ + $\frac{1}{5}$ – $\frac{1}{7}$ + $\frac{1}{7}$ – $\frac{1}{9}$ + $\frac{1}{9}$ – $\frac{1}{11}$
    ⇒$\frac{1}{2}$.(1 – $\frac{1}{11}$)
    ⇒$\frac{1}{2}$.$\frac{10}{11}$ = $\frac{5}{11}$  

    d) 2/1.3+2/3.5+ 2/5.7+ 2/7.9+ 2/9.11

    <=>1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11

    => 1 – 1/11 = 10/11

    #Chucbanhoctot

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  2. Đáp án:

    c,`1/(1.3)+1/(3.5)+1/(5.7)+1/(7.9)+1/(9.11)`

    `=1/2. (2/(1.3)+2/(3.5)+2/(5.7)+2/(7.9)+2/(9.11))`

    `=1/2.(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)`

    `=1/2.(1-1/11)`

    `=1/2 . 10/11=5/11`

    d,`2/(1.3)+2/(3.5)+2/(5.7)+2/(7.9)+2/(9.11)`

    `=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11`

    `=1-1/11`

    `=10/11`

    e,`(1+1/2).(1+ 1/3). ( 1+1/4)…(1+ 1/100)`

    `=3/2 . 4/3 . 5/4 ……………. 101/100`

    `=(3.4.5….101)/(2.3.4…..100)`

    `=101/2`

    f/ `( 1-1/2). ( 1-1/3).( 1-1/4) …(1-1/100)`

    `=1/2 . 2/3 . 3/4 ……….. 99/100`

    `=(1.2.3…99)/(2.3.4….100)`

    `=1/100`

    XIN HAY NHẤT NHA

    @hoang

     

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