2: (3x-4)^2+2>=2

=>5/(3x-4)^2+2<=5/2

=>B>=-5/2

Dấu = xảy ra khi x=4/3

4: D=(3x^2+7-4)/(3x^2+7)=1-4/3x^2+7

3x^2+7>=7

=>4/3x^2+7<=4/7

=>-4/3x^2+7>=-4/7

=>D>=3/7

Dấu = xảy ra khi x=0

2) B = \(\dfrac{-5}{\left(3x-4\right)^2+2}\) 

Ta có: ( 3x-4)² \(\ge\) 0 , \(\forall\) x

=> ( 3x-4)² +2 \(\ge\) 2, \(\forall\) x

=> \(\dfrac{1}{\left(3x-4\right)^2+2}\) \(\le\) \(\dfrac{1}{2}\) , \(\forall\) x

=> \(\dfrac{-5}{\left(3x-4\right)^2+2}\) \(\ge\) \(\dfrac{-5}{2}\) , \(\forall\) x

=> B \(\ge\) \(\dfrac{-5}{2}\) 

Vậy B đạt GTNN khi bằng \(\dfrac{-5}{2}\) 

Dấu "= " xảy ra khi 3x - 4 = 0

4) D=\(\dfrac{3x^2+3}{3x^2+7}\) 

= 1 - \(\dfrac{4}{3x^2+7}\) 

Ta có: 3x² \(\ge\) 0, \(\forall\) x

=> 3x² +7 \(\ge\) 7, \(\forall\) x

=> \(\dfrac{1}{3x^2+7}\) \(\le\) \(\dfrac{1}{7}\) 

=> \(\dfrac{4}{3x^2+7}\) \(\le\) \(\dfrac{4}{7}\) 

=> 1 - \(\dfrac{4}{3x^2+7}\) \(\ge\) \(\dfrac{3}{7}\) 

Vậy D đạt GTNN khi bằng \(\dfrac{3}{7}\) 

Dấu "=" xảy ra khi x = 0

\(C=x^2+2x+1\dfrac{1}{2}\\ \Rightarrow C=\left(x^2+2x+1\right)+\dfrac{1}{2}\\ \Rightarrow C=\left(x+1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-1\)

Vậy \(C_{min}=\dfrac{1}{2}\Leftrightarrow x=-1\)

 \(C=x^2+2x+1\dfrac{1}{2}.\\ C=x^2+2x+1+\dfrac{1}{2}.\\ C=\left(x+1\right)^2+\dfrac{1}{2}.\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\in R.\\ \dfrac{1}{2}>0. \)

\(\Rightarrow\left(x+1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}.\)

Dấu "=" xảy ra khi \(x+1=0.\Leftrightarrow x=-1.\)

Vậy GTNN của biểu thức C là \(\dfrac{1}{2}\) khi x = -1.

x = 1

x^2 - 4x + 1 = x^2 - 4x + 4 - 3 = ( x- 2 )^2 - 3 

Vậy GTnn là 3 khi x = 2