x-1/2014+x-2/2013+x-3/2012=x-10/2005=+x-11/2004+x-12/2003 05/12/2021 Bởi Vivian x-1/2014+x-2/2013+x-3/2012=x-10/2005=+x-11/2004+x-12/2003
Đáp án : `x=2015` Giải thích các bước giải : `(x-1)/(2014)+(x-2)/(2013)+(x-3)/(2012)=(x-10)/(2005)+(x-11)/(2004)+(x-12)/(2003)` `<=>(x-1)/(2014)+(x-2)/(2013)+(x-3)/(2012)-3=(x-10)/(2005)+(x-11)/(2004)+(x-12)/(2003)-3` `<=>((x-1)/(2014)-1)+((x-2)/(2013)-1)+((x-3)/(2012)-1)=((x-10)/(2005)-1)+((x-11)/(2004)-1)+((x-12)/(2003)-1)` `<=>(x-1-2014)/(2014)+(x-2-2013)/(2013)+(x-3-2012)/(2012)=(x-10-2005)/(2005)+(x-11-2004)/(2004)+(x-12-2003)/(2003)` `<=>((x-1-2014)/(2014)+(x-2-2013)/(2013)+(x-3-2012)/(2012))-((x-10-2005)/(2005)+(x-11-2004)/(2004)+(x-12-2003)/(2003))=0` `<=>(x-2015)/(2014)+(x-2015)/(2013)+(x-2015)/(2012)-(x-2015)/(2005)-(x-2015)/(2004)-(x-2015)/(2003)=0` `<=>(x-2015)(1/(2014)+1/(2013)+1/(2012)-1/(2005)-1/(2004)-1/(2003))=0` Vì `1/(2014)+1/(2013)+1/(2012)-1/(2005)-1/(2004)-1/(2003) ≠ 0` `=>x-2015=0` `<=>x=2015` Vậy `x=2015` ~Chúc bạn học tốt !!!~ Bình luận
Đáp án: `(x-1)/2014+(x-2)/2013+(x-3)/2012=(x-10)/2005+(x-11)/2004+(x-12)/2003` `-> ((x-1)/2014-1)+((x-2)/2013-1)+((x-3)/2012-1)=((x-10)/2005-1)+((x-11)/2004-1)+((x-12)/2003-1)` `-> (x-2015)/2014+(x-2015)/2013+(x-2015)/2012=(x-2015)/2005+(x-2015)/2004+(x-2015)/2003` `-> (x-2015)/2014+(x-2015)/2013+(x-2015)/2012-(x-2015)/2005-(x-2015)/2004-(x-2015)/2003=0` `-> (x-2015).(1/2014+1/2013+1/2012-1/2005-1/2004-1/2003)=0` Mà `(1/2014+1/2013+1/2012-1/2005-1/2004-1/2003)\ne0` `-> x-2015=0` `-> x=2015` Bình luận
Đáp án :
`x=2015`
Giải thích các bước giải :
`(x-1)/(2014)+(x-2)/(2013)+(x-3)/(2012)=(x-10)/(2005)+(x-11)/(2004)+(x-12)/(2003)`
`<=>(x-1)/(2014)+(x-2)/(2013)+(x-3)/(2012)-3=(x-10)/(2005)+(x-11)/(2004)+(x-12)/(2003)-3`
`<=>((x-1)/(2014)-1)+((x-2)/(2013)-1)+((x-3)/(2012)-1)=((x-10)/(2005)-1)+((x-11)/(2004)-1)+((x-12)/(2003)-1)`
`<=>(x-1-2014)/(2014)+(x-2-2013)/(2013)+(x-3-2012)/(2012)=(x-10-2005)/(2005)+(x-11-2004)/(2004)+(x-12-2003)/(2003)`
`<=>((x-1-2014)/(2014)+(x-2-2013)/(2013)+(x-3-2012)/(2012))-((x-10-2005)/(2005)+(x-11-2004)/(2004)+(x-12-2003)/(2003))=0`
`<=>(x-2015)/(2014)+(x-2015)/(2013)+(x-2015)/(2012)-(x-2015)/(2005)-(x-2015)/(2004)-(x-2015)/(2003)=0`
`<=>(x-2015)(1/(2014)+1/(2013)+1/(2012)-1/(2005)-1/(2004)-1/(2003))=0`
Vì `1/(2014)+1/(2013)+1/(2012)-1/(2005)-1/(2004)-1/(2003) ≠ 0`
`=>x-2015=0`
`<=>x=2015`
Vậy `x=2015`
~Chúc bạn học tốt !!!~
Đáp án:
`(x-1)/2014+(x-2)/2013+(x-3)/2012=(x-10)/2005+(x-11)/2004+(x-12)/2003`
`-> ((x-1)/2014-1)+((x-2)/2013-1)+((x-3)/2012-1)=((x-10)/2005-1)+((x-11)/2004-1)+((x-12)/2003-1)`
`-> (x-2015)/2014+(x-2015)/2013+(x-2015)/2012=(x-2015)/2005+(x-2015)/2004+(x-2015)/2003`
`-> (x-2015)/2014+(x-2015)/2013+(x-2015)/2012-(x-2015)/2005-(x-2015)/2004-(x-2015)/2003=0`
`-> (x-2015).(1/2014+1/2013+1/2012-1/2005-1/2004-1/2003)=0`
Mà `(1/2014+1/2013+1/2012-1/2005-1/2004-1/2003)\ne0`
`-> x-2015=0`
`-> x=2015`