Tìm x biết:(x-1)/2011+(x-2)/2010+(x-3)/2009=3 12/08/2021 Bởi Daisy Tìm x biết:(x-1)/2011+(x-2)/2010+(x-3)/2009=3
Đáp án: x=2012 Giải thích các bước giải: $\eqalign{ & \frac{{x – 1}}{{2011}} + \frac{{x – 2}}{{2010}} + \frac{{x – 3}}{{2009}} = 3 \cr & \Leftrightarrow (\frac{{x – 1}}{{2011}} – 1) + (\frac{{x – 2}}{{2010}} – 1) + (\frac{{x – 3}}{{2009}} – 1) = 0 \cr & \Leftrightarrow \frac{{x – 2012}}{{2011}} + \frac{{x – 2012}}{{2010}} + \frac{{x – 2012}}{{2009}} = 0 \cr & \Leftrightarrow (x – 2012)(\frac{1}{{2011}} + \frac{1}{{2010}} + \frac{1}{{2009}}) = 0 \cr & \Leftrightarrow x – 2012 = 0(do\,\frac{1}{{2011}} + \frac{1}{{2010}} + \frac{1}{{2009}} \ne 0) \cr & \Leftrightarrow x = 2012 \cr} $ Bình luận
Đáp án: x=2012
Giải thích các bước giải:
$\eqalign{ & \frac{{x – 1}}{{2011}} + \frac{{x – 2}}{{2010}} + \frac{{x – 3}}{{2009}} = 3 \cr & \Leftrightarrow (\frac{{x – 1}}{{2011}} – 1) + (\frac{{x – 2}}{{2010}} – 1) + (\frac{{x – 3}}{{2009}} – 1) = 0 \cr & \Leftrightarrow \frac{{x – 2012}}{{2011}} + \frac{{x – 2012}}{{2010}} + \frac{{x – 2012}}{{2009}} = 0 \cr & \Leftrightarrow (x – 2012)(\frac{1}{{2011}} + \frac{1}{{2010}} + \frac{1}{{2009}}) = 0 \cr & \Leftrightarrow x – 2012 = 0(do\,\frac{1}{{2011}} + \frac{1}{{2010}} + \frac{1}{{2009}} \ne 0) \cr & \Leftrightarrow x = 2012 \cr} $