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\(A=\left|2014-x\right|+\left|2015-x\right|+2016-x\)
Ta xét 4 trường hợp xảy ra:

TH1: \(x< 2014\)

\(A=2014-x+2015-x+2016-x\)

\(=6045-3x>3\) ( Vì \(x< 2014\) ) (1)

TH2: \(2014\le x\le2015\)

\(A=x-2014+2015-x+2016-x\)

\(=2017-x>2\) ( Vì \(x< 2015\) ) (2)

TH3: \(2015\le x< 2016\)

\(A=x-2014+x-2015+2016-x\)

\(=x-2013\ge2\) ( Vì \(x\ge2015\) ) (3)

TH4: \(x< 2016\)

\(A=x-2014+x-2015+x-2016\)

\(=3x-6045>3\) ( Vì \(x>2016\) ) (4)

Từ (1), (2), (3) và (4) \(\Rightarrow A\ge2\)

Vậy A nhỏ nhất =2 khi x=2015.

\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)

\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)

\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)

\(M=343-8x^3-64+8x^3\)

\(M=279\)

Vậy M có giá trị 279 với mọi x

\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)

\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)

\(P=16x^3-8x^2+4x-2\)

Thay \(x=10\) vào P ta có:

\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)

Vậy P có giá trị 15238 tại x=10

a: M=343-8x^3-64+8x^3=279

b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3

=16x^3-8x^2+4x-2

=16*10^3-8*10^2+4*10-2=15238

\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)

\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)

\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)

\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)

Ta có :

\(N=\left|x-2014\right|+\left|2015-x\right|\ge\left|x-2014+2015-x\right|=1\)

Dấu "=" xảy ra khi :

\(\Leftrightarrow\left(x-2014\right)\left(2015-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2014\ge0\\2015-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2014\le0\\2015-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2014\\2015\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2014\\2015\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2014\le x\le2015\\x\in\varnothing\end{matrix}\right.\)

Vậy \(N_{Min}=1\Leftrightarrow2014\le x\le2015\)

N= | x-2014 | +|2015 -x| ≥| x-2014 + 2015 -x | = | 1| = 1

dấu "=" khi x-2014 = 2015 - x

<=> x = 2014,5

vậy gtnn N = 1 khi x = 2014,5

a) Có \(\left(x-1\right)^2\ge0\)

<=> A \(\ge2014\)

Dấu "=" <=> x = 1

b) Có \(\left|x+4\right|\ge0\)

<=> B \(\ge2014\)

Dấu "=" <=> x = -4

a) \(A=\left(x-1\right)^2+2014\ge2014\)

Dấu = xảy ra khi x = 1

b) \(B=\left|x+4\right|+2014\ge2014\)

Dấu = xảy ra khi x = -4