B=2(x^2+2.x.1/4 +1/16)^2 -57/8

  =2.(x+1/4)^2 -57/8

MinB=-57/8 khi x=-1/4

\(B=-14+2x^2+x=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{113}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{113}{8}\ge-\dfrac{113}{8}\)\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)

\(B=2x^2+x-14\)

\(=2\left(x^2+\dfrac{1}{2}x-7\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{113}{16}\right)\)

\(=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{113}{8}\ge-\dfrac{113}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)

`B=x^2 +y^2 -2x+4y+2010`

`=x^2 -2x+1+y^2 +4y+4+2005`

`=(x-1)^2 + (y+2)^2 +2005 >= 2005`

Dấu "=" xảy ra `<=>{(x-1=0),(y+2=0):}<=>{(x=1),(y=-2):}`

Vậy `B_(min) = 2005 <=> {(x=1),(y=-2):}`

`B=x^2+y^2-2x+4y+2010`

`B=x^2-2x+y^2+4y+2010`

`B= x^2-2.x.1+1^2-1^2 +y^2+2y.2+2^2-2^2+2010`

`B= (x^2-2x+1)+(y^2+4y+4)-1-4+2010`

`B= (x-1)^2 +(y+2)^2 +2005≥2005`

nên `B` đạt GTNN là `B=2005`

khi đó \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\) `<=>`\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

b: \(=\dfrac{2014\cdot2015^2+2014\cdot2016-2016\cdot2015^2+2016\cdot2014}{2014\cdot2013^2-2014\cdot2012-2012\cdot2013^2-2012\cdot2014}\)

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

\(=\dfrac{\left(-2\right)\cdot\left(2015^2-2015^2+1\right)}{\left(-2\right)\cdot\left(2013^2+2013^2-1\right)}=\dfrac{1}{2\cdot2013^2}\)

a,|2x+2014| lớn hơn hoặc bằng 0

   |x+2015| lớn hơn hoặc bằng 0

mà |2x+2014|+|x+2015|=0

=> |2x+2014|=0

       x=-1007

|x+2015|=0

  x=-2015 (vô lí)

=> x thuộc tập hp rỗng

\(A=\left|2x-1\right|+5\ge5\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(B=\left|2x-2014\right|+2015\ge2015\forall x\)

Dấu '=' xảy ra khi x=1007

\(x-2014-\frac{2015}{2013}+x-2013-\frac{2015}{2014}+x-2014-\frac{2013}{2015}=3\)

\(\Rightarrow\left(x+x+x\right)+\left(-2014-2014\right)-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)

\(3x-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)

\(3x=3+2013+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2013}{2015}\)

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