Question

Answer 1

      \(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}\)≥\(\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
 ⇒(\(\dfrac{x-3}{2011}-1\))+(\(\dfrac{x-2}{2012}\)-1)-(\(\dfrac{x-2012}{2}-1\))-(\(\dfrac{x-2011}{3}-1\))≥0
⇒\(\dfrac{x-3-2011}{2011}+\dfrac{x-2-2012}{2012}-\dfrac{x-2012-2}{2}-\dfrac{x-2011-3}{3}\)≥0
⇒\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}\)≥0
⇒(x-2014)(\(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\))≥0
Ta thấy rằng: (\(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\))≠0( ∀x)
⇒x-2014≥0⇒x≥2014
Vậy bất phương trình có tập nghiệm là S={x/x≥2014}

Answer 2

Giúp em với