Câu hỏi của Feliks Zemdegs

Question

Tìm x biết rằng (2015x−2014)3=8(x−1)3+(2013x−2012)3.

Trần Thị Loan · Accepted Answer

PT <=> (2015x - 2014)3 = (2x - 2)3 + (2013x - 2012)3<=> (2015x - 2014)3 = (2x - 2 + 2013x - 2012). [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]<=> (2015x - 2014)3 = (2015x - 2014). [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]<=> (2015x - 2014).[ (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]] = 0 <=> 2015.x - 2014 = 0 hoặc (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2] = 0 +) 2015x - 2014 = 0 => x = 2014/2015+) (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2] = 0<=> [(2x - 2) + (2013x - 2012)]2 - (2x - 2)2 + (2x - 2).(2013x - 2012) - (2013x - 2012)2 = 0 <=> 3. (2x - 2).(2013x - 2012) = 0 <=> 2x - 2 = 0 hoặc 2013x - 2012 = 0 <=> x = 1 hoặc x = 2012/2013Vậy....Câu trả lời: PT <=> (2015x - 2014)3 = (2x - 2)3 + (2013x - 2012)3<=> (2015x - 2014)3 = (2x - 2 + 2013x - 2012). [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]<=> (2015x - 2014)3 = (2015x - 2014). [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]<=> (2015x - 2014).[ (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2]] = 0 <=> 2015.x - 2014 = 0 hoặc (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2] = 0 +) 2015x - 2014 = 0 => x = 2014/2015+) (2015x - 2014)2 - [(2x-2)2 - (2x - 2).(2013x - 2012) + (2013x - 2012)2] = 0<=> [(2x - 2) + (2013x - 2012)]2 - (2x - 2)2 + (2x - 2).(2013x - 2012) - (2013x - 2012)2 = 0 <=> 3. (2x - 2).(2013x - 2012) = 0 <=> 2x - 2 = 0 hoặc 2013x - 2012 = 0 <=> x = 1 hoặc x = 2012/2013Vậy....

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)