\( \dfrac{x+2001}{2016}+\dfrac{x+2002}{2017}>\dfrac{x+2003}{2018}+\dfrac{x+2004}{2019}\\↔(\dfrac{x+2001}{2016}-1)+(\dfrac{x+2002}{2017}-1)>(\dfrac{x+2003}{2018}-1)+(\dfrac{x+2004}{2019}-1)\\↔\dfrac{x-15}{2016}+\dfrac{x-15}{2017}>\dfrac{x-15}{2018}+\dfrac{x-15}{2019}\\↔\dfrac{x-15}{2016}+\dfrac{x-15}{2017}-\dfrac{x-15}{2018}-\dfrac{x-15}{2019}>0\\↔(x-15)(\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019})>0\\Vì\,\,\dfrac{1}{2016}>\dfrac{1}{2018}\\\dfrac{1}{2017}>\dfrac{1}{2019}\\→\dfrac{1}{2016}+\dfrac{1}{2017}>\dfrac{1}{2018}+\dfrac{1}{2019}\\↔\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}>0\\→x-15>0\\↔x>15\\\text{Vậy tập nghiệm của bất phương trình là}\,\,\{x|x>15\}\)
Đáp án + Giải thích các bước giải:
`(x+2001)/(2016)+(x+2002)/(2017)>(x+2003)/(2018)+(x+2004)/(2019)`
`<=>((x+2001)/(2016)-1)+((x+2002)/(2017)-1)>((x+2003)/(2018)-1)+((x+2004)/(2019)-1)`
`<=>((x+2001)/(2016)-(2016)/(2016))+((x+2002)/(2017)-(2017)/(2017))>((x+2003)/(2018)-(2018)/(2018))+((x+2004)/(2019)-(2019)/(2019))`
`<=>(x-15)/(2016)+(x-15)/(2017)>(x-15)/(2018)+(x-15)/(2019)`
`<=>(x-15)/(2016)+(x-15)/(2017)-(x-15)/(2018)-(x-15)/(2019)>0`
`<=>(x-15)((1)/(2016)+(1)/(2017)-(1)/(2018)-(1)/(2019))>0`
Vì `(1)/(2016)+(1)/(2017)-(1)/(2018)-(1)/(2019)>0`
`=>x-15>0`
`<=>x>15`
Vậy nghiệm bất phương trình là : `x>15`
\( \dfrac{x+2001}{2016}+\dfrac{x+2002}{2017}>\dfrac{x+2003}{2018}+\dfrac{x+2004}{2019}\\↔(\dfrac{x+2001}{2016}-1)+(\dfrac{x+2002}{2017}-1)>(\dfrac{x+2003}{2018}-1)+(\dfrac{x+2004}{2019}-1)\\↔\dfrac{x-15}{2016}+\dfrac{x-15}{2017}>\dfrac{x-15}{2018}+\dfrac{x-15}{2019}\\↔\dfrac{x-15}{2016}+\dfrac{x-15}{2017}-\dfrac{x-15}{2018}-\dfrac{x-15}{2019}>0\\↔(x-15)(\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019})>0\\Vì\,\,\dfrac{1}{2016}>\dfrac{1}{2018}\\\dfrac{1}{2017}>\dfrac{1}{2019}\\→\dfrac{1}{2016}+\dfrac{1}{2017}>\dfrac{1}{2018}+\dfrac{1}{2019}\\↔\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}>0\\→x-15>0\\↔x>15\\\text{Vậy tập nghiệm của bất phương trình là}\,\,\{x|x>15\}\)