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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: 1. Tính: B=( 1-$\frac{1}{2}$ ).(1-$\frac{1}{3}$ ).( 1-$\frac{1}{4}$ )-(1-$\frac{n}{n+1}$ ) Với n ∈ N 2.Tìm x biết: $\frac{x+3}{200

Home/ Toán Lớp 7: 1. Tính: B=( 1-$\frac{1}{2}$ ).(1-$\frac{1}{3}$ ).( 1-$\frac{1}{4}$ )-(1-$\frac{n}{n+1}$ ) Với n ∈ N 2.Tìm x biết: $\frac{x+3}{200

Toán Lớp 7: 1. Tính: B=( 1-$\frac{1}{2}$ ).(1-$\frac{1}{3}$ ).( 1-$\frac{1}{4}$ )-(1-$\frac{n}{n+1}$ ) Với n ∈ N 2.Tìm x biết: $\frac{x+3}{200

Tháng Tám 10, 2022 Toán học 1 Comment 0 views

Toán Lớp 7: 1. Tính:
B=( 1-$\frac{1}{2}$ ).(1-$\frac{1}{3}$ ).( 1-$\frac{1}{4}$ )….(1-$\frac{n}{n+1}$ )
Với n ∈ N
2.Tìm x biết:
$\frac{x+3}{2007}$ – $\frac{x+3}{2008}$ = $\frac{x+3}{2010}$ – $\frac{x+3}{2009}$

Comments ( 1 )

  1. dinhcongson
    dinhcongson
    10 Tháng Tám, 2022 at 10:21 chiều
    Reply
    $2.$
    $\frac{x+3}{2007}$ – $\frac{x+3}{2008}$ = $\frac{x+3}{2010}$ – $\frac{x+3}{2009}$ 
    $⇒$ $\frac{x+3}{2007}$ – $\frac{x+3}{2008}$ – $\frac{x+3}{2010}$ + $\frac{x+3}{2009}$=$0$
    $⇔$ ($x+3$).$\frac{1}{2007}$ – $\frac{1}{2008}$ – $\frac{1}{2010}$ + $\frac{1}{2009}$ = $0$
    $⇔$ ($x+3=0$) (do $\frac{1}{2007}$ – $\frac{1}{2008}$ – $\frac{1}{2010}$ + $\frac{1}{2009}$ > $0$
    $⇒$ $x=-3$
    $\text{Bài 1}$:

    toan-lop-7-1-tinh-b-1-frac-1-2-1-frac-1-3-1-frac-1-4-1-frac-n-n-1-voi-n-n-2-tim-biet-frac-3-200

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222-9+11+12:2*14+14 = ? ( )

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