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A = |x + 1| + | x + 2| + |x + 3| + ............... + |x + 2016| + 100
Đặt : A' = |x + 1| + | x + 2| + |x + 3| + ............... + |x + 2016|
=> A' = |x + 1| + |-x - 2| + |x + 3| + ............... + |-x - 2016|
Áp dụng BĐT |a| + |b| \(\ge\) |a + b| , có :
|x + 1| + |-x - 2| + |x + 3| + ............... + |-x - 2016| \(\ge\) |x + 1 - x - 2 + x + 3 - x - 4 + ....... + x + 2015 - x - 2016|
<=> A' \(\ge\) |1 - 2 + 3 - 4 + ......... + 2015 - 2016| = |-1008| = 1008
=> A \(\ge\) 1008 + 100 = 1108
=> MinA = 1108
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x+(x+1)+(x+2)+(x+3)+...+(x+2016)=2017
( x + x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 2016 ) = 2017
2017x + 2033136 = 2017
2017x = 2017 - 2033136
2017x = -2031119
x = -2031119 : 2017
x = -1007
Ta có : x + (x + 1) + (x + 2) + (x + 3) +......+ (x + 2016) = 2017
=> (x + x + x + ..... + x) + (1 + 2 + 3 + .... + 2016) = 2017
=> 2017x + 2033136 = 2017
=> 2017x = 2017 - 2033136
=> 2017x = -203119
=> x = -203119 : 2017
=> x = -1007
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vi x+2016> hoac bang 0 va x+2017>hoac =0=>gtnn cua hang thuc tren =0
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a: ĐKXĐ: x<>1; x<>2; x<>-2; x<>-1
\(P=\dfrac{2017x+2017-2016x+2016-2014x-2016}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2015x+2017}{x^2-4}\)
- Ta có: \(\left|x+1\right|\ge0\forall x\)
\(\left|x+2\ge0\forall x\right|\)
......
\(\left|x-2017\ge0\forall x\right|\)
- Suy ra: \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+.....+\left|x+2017\right|\ge0\forall x\)
=> \(\left|x+1\right|+\left|x+2\right|+....+\left|x+2017\right|+100\ge100\forall x\)
- Dấu bằng xảy ra khi
\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2017\right|=0\)
- Suy ra : Giá trị nhỏ nhất của A ( MinA) = 100
<=> \(\left|x+1\right|+\left|x+2\right|+...+\left|x+2017\right|=0\).