Violympic toán 7

Monkey D Luffy

Tìm x :

\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)

Nguyễn Thanh Hằng
18 tháng 1 2018 lúc 21:09

\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)

\(\Leftrightarrow\dfrac{x-1}{2011}+\dfrac{x-2}{2012}+\dfrac{x-3}{2009}-\dfrac{x-4}{2008}=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)+\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-4}{2008}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)

\(\Leftrightarrow\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

\(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\)

\(\Leftrightarrow x-2012=0\Leftrightarrow x=2012\)

Vậy ...

Bình luận (0)
Shizadon
18 tháng 1 2018 lúc 21:17

\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)

=> \(\dfrac{x-1}{2011}-1+\dfrac{x-2}{2010}-1+\dfrac{x-3}{2009}-1=\dfrac{x-4}{2008}-1-2\)

=>\(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}=\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}\)

=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}=0\)=> x - 2012 ( \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\)) = 0

\(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\) \(\ge\) 0

=> x - 2012 = 0

=> x = 2012

Bình luận (0)

Các câu hỏi tương tự
nguyễn ly
Xem chi tiết
Tưởng Y Y
Xem chi tiết
Huyền Anh Lê
Xem chi tiết
Thuy Khuat
Xem chi tiết
Nguyễn Đăng Khoa
Xem chi tiết
dream XD
Xem chi tiết
Aquarius
Xem chi tiết
Trần Hoài khánh Trang
Xem chi tiết
Lê Vũ Anh Thư
Xem chi tiết