(x+1/2013) + (x+2/2012) + (x+3/2011) = (x+4/2010) + (x+5/2009) + (x+6/2008) help me!!!

0 bình luận về “(x+1/2013) + (x+2/2012) + (x+3/2011) = (x+4/2010) + (x+5/2009) + (x+6/2008) help me!!!”

1. `(x+1)/2013 + (x+2)/2012 + (x+3)/2011 = (x+4)/2010 + (x+5)/2009 + (x+6)/2008`

   `⇒ (x+1)/2013 + (x+2)/2012 + (x+3)/2011 – (x+4)/2010 – (x+5)/2009 – (x+6)/2008 = 0`

   `⇒ ((x+1)/2013 + 1) + ((x+2)/2012 + 1) + ((x+3)/2011 + 1) – ((x+4)/2010 + 1) – ((x+5)/2009 + 1) – ((x+6)/2008 + 1) = 0` 

   `⇒ (x+2014)/2013 + (x+2014)/2012 + (x+2014)/2011 – (x+2014)/2010 – (x+2014)/2009 – (x+2014)/2008 = 0`

   `⇒ (x + 2014). (1/2013 + 1/2021 + 1/2011 – 1/2010 – 1/2009 – 1/2008) = 0`

   Mà `(1/2013 + 1/2021 + 1/2011 – 1/2010 – 1/2009 – 1/2008) \ne 0`

   `⇒ x + 2014 = 0`

   `⇒ x = -2014`

   Bình luận

2. Đáp án:

   `(x+1)/2013+(x+2)/2012+(x+3)/2011=(x+4)/2010+(x+5)/2009+(x+6)/2008`

   `<=> (x+1)/2013 +1 +(x+2)/2012 +1 +(x+3)/2011 +1 =(x+4)/2010 +1 +(x+5)/2009 +1 +(x+6)/2008 +1 `

   `<=> (x+2014)/2013+(x+2014)/2012+(x+2014)/2011=(x+2014)/2010+(x+2014)/2009+(x+2014)/2008`

   `<=> (x+2014)/2013+(x+2014)/2012+(x+2014)/2011-(x+2014)/2010-(x+2014)/2009-(x+2014)/2008=0`

   `<=> (x+2014).(1/2013+1/2012+1/2011-1/2010-1/2009-1/2008)=0`

   Mà `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`

   `=> x+2014=0`

   `<=> x=-2014`

   Vậy `x=-2014`

   Bình luận