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Ta có: \(112⋮x\)
\(140⋮x\)
Do đó: \(x\inƯC\left(112;140\right)\)
\(\Leftrightarrow x\in\left\{1;2;4;7;14;28\right\}\)
mà 10<x<20
nên x=14
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt vế trái là A ta có:
\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow A=\frac{x-1}{x+1}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\Leftrightarrow x=2003\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)
<=> \(\frac{1}{x+1}=\frac{1}{2021}\)
<=> x + 1 = 2021
<=> x = 2020
Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}:2\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}:2=\frac{1}{2001}\Rightarrow x+1=2001\Rightarrow x=2000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: 1/3 + 1/6 + ... + 2/x(x+1) = 2/2.3 + 2/3.4 +.......2/x(x+1) = 2(1/2.3 +1/3.4 +.....+1/x(x+1)) = 2.(1/2-1/3+1/3-1/4+....+1/x-1/(x+1))= 2.(1/2-1/(x+1)) = 1-2/(x+1)
giải 1-2/(x+1) = 2007/2009 ta được x=2008
![](https://rs.olm.vn/images/avt/0.png?1311)
= 2/(2.3) + 2/3.4 + 2/4.5 +...+ 2/x(x+1)
= 2 [1/2-1/3+1/3-1/4+...+1/x-1/(x+1)]
=2[1/2-1/(x+1)]= (x-1)/(x+1)
= 2001/2003
==> x=2002
Tìm số tự nhiên x, biết rằng 162 ⋮ x; 360 ⋮ x và 10 < x < 20 :
A. x = 6
B. x = 9
C. x = 18
D. x = 36
![](https://rs.olm.vn/images/avt/0.png?1311)
Đáp án là C
Vì 162 ⋮ x; 360 ⋮ x ⇒ x ∈ ƯC (162, 360)
Ta có:
162 = 2. 3 4
360 = 2 3 . 3 2 .5
Suy ra, ƯCLN ( 162, 360) = 2. 3 2 = 18
ƯC ( 162, 360) = Ư (18) = {1, 2, 3, 6, 9, 18} ⇒ x ∈ {1, 2, 3, 6, 9, 18}
Vì 10 < x < 20 ⇒ x = 18
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 112 ⋮ x ; 140 ⋮ x nên x ∈ ƯC(112, 140).
Ta có 112 = 24.7 ; 140 = 22.5.7
⇒ ƯCLN(112, 140) = 22.7 = 28.
⇒ ƯC(112, 140) = Ư(28) = {1, 2, 4, 7, 14, 28}.
⇒ x ∈ {1; 2; 4; 7; 14; 28}.
Mà 10 < x < 20 nên x = 14.
2 ( 118 - x ) + 10 = 10
2 ( 118 - x ) = 10 - 10
2 ( 118 - x ) = 0
118 - x = 0÷ 2
118 - x = 0
x = 118 - 0
x = 118
Vậy x = 118
2 ( 118 - x ) + 10 = 10
2 ( 118 - x ) = 10 - 10
2 ( 118 - x ) = 0
118 - x = 0 : 2
118 - x = 0
x = 118 - 0
x = 118
Vậy x = 118