Tìm x biết rằng (2015x−2014)3=8(x−1)3+(2013x−2012)3.
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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 = 2012/2013
Vậy....
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(2015x - 2014)3 = 8(x - 1)3 + (2013x - 2012)3
<=> 6(x - 1)(2013x - 2012)(2015x - 2014) = 0
Tới đây thì xong rồi
|x+1| + |x+2| + |x+3| + .......... + |x+2014| = 2015x
Ta có :
|x+1| \(\ge\)0
|x+2| \(\ge\)0
|x+3| \(\ge\)0
..........
|x+2014| \(\ge\)0
=> |x+1| + |x+2| + |x+3| +..........+ |x+2014| \(\ge\)0
=> 2015x \(\ge\)0
Mà 2015 \(\ge\)0
=> x \(\ge\)0
=> |x+1| + |x+2| + |x+3| +..........+ |x+2014|
= x + 1 + x + 2 + x + 3 +.................... + x + 2014 = 2015x
=> 2014x + (1 + 2 + 3 +............ + 2014) = 2015x
=> 1 + 2 + 3 + 4 + ........................ + 2014 = x
=> x = 2029105