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

\(A_{min}=-1\) khi \(x=-2\)

\(A=\dfrac{4x^2+4-4x^2+4x-1}{x^2+1}=4-\dfrac{\left(2x-1\right)^2}{x^2+1}\le4\)

\(A_{max}=4\) khi \(x=\dfrac{1}{2}\)

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

\(A_{min}=-1\) khi \(x=2\)

\(A=\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}=4-\dfrac{\left(2x+1\right)^2}{x^2+1}\le4\)

\(A_{max}=4\) khi \(x=-\dfrac{1}{2}\)

GTNN:

Vì \(\sqrt{x}\ge0\Rightarrow3\sqrt{x}\ge0\Rightarrow P=3\sqrt{x}+1\ge1\)

Dấu bằng xảy ra <=> x=0

\(Q=\sqrt{x}+\dfrac{25}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{25}{\sqrt{x}}}=10\)

Dấu = xảy ra khi x=25

Ta có: \(A=\left|x-2013\right|+\left|x-1\right|=\left|2013-x\right|+\left|x-1\right|\ge\left|2013-x+x-1\right|=2012\)

Dấu "=" xảy ra khi \(\left(2013-x\right)\left(x-1\right)\ge0\Leftrightarrow1\le x\le2013\)

Vậy GTNN của A = 2012 khi 1 =< x =< 2013

`A=-10/(sqrtx+5)(x>=0)`

`x>=0=>sqrtx>=0`

`=>sqrtx+5>=5>0`

`=>10/(sqrtx+5)<=10/5=2`

`=>A>=-2`

Dấu "=" xảy ra khi `x=0`

Vậy GTNN `A=-2<=>x=0`

\(S=\dfrac{2018x^2-2.2018x+2018^2}{2018x^2}=\dfrac{2017x^2+x^2-2.2018x+2018^2}{2018x^2}=\dfrac{2017}{2018}+\dfrac{\left(x-2018\right)^2}{x^2}\ge\dfrac{2017}{2018}\)

\(S_{min}=\dfrac{2017}{2018}\) khi \(x=2018\)

\(2x^2+4x+15=2.\left(x^2+2x+1\right)+13=2.\left(x+1\right)^2+13\ge13,\forall x\inℝ\\ \)

Dấu "=" xảy ra <=> x=-1

Vậy \(Min\left(A\right)=13\Leftrightarrow x=-1\)

2x²+4x+15

=2(x²+2x+1)+13

=2(x+1)²+13

Có 2(x+1)²\(\ge\)0 \(\forall x\in R\)

=>2(x+1)²+13\(\ge13\forall x\in R\)

Vậy GTNN của phương trình trên là 13

\(2x^2+4x+15=2\left(x^2+2x+\frac{15}{2}\right)\)

\(=2\left(x^2+2x+1+\frac{13}{2}\right)\)

\(=2\left(x+1\right)^2+13\)

Vì:\(2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x+1\right)^2+13\ge13\forall x\)

Dấu = xảy ra khi \(2\left(x+1\right)^2=0\Rightarrow x=-1\)

vậy gtnn của bt là 13 tại x=-1