Ta có: \(D=\frac{3x^2+2x+1}{2x^2+1}\)

\(=\frac{3x^2+1,5+2x-0,5}{2x^2+1}=\frac{1,5\left(2x^2+1\right)+2x-0,5}{2x^2+1}\)

\(=1,5+\frac{2x-0,5}{2x^2+1}\)

Đặt \(A=\frac{2x-0,5}{2x^2+1}\)

=>\(A\left(2x^2+1\right)=2x-0,5\)

=>\(2A\cdot x^2+A-2x+0,5=0\)

=>\(2A\cdot x^2-2x+A+0,5=0\) (1)

\(\Delta=\left(-2\right)^2-4\cdot2A\cdot\left(A+0,5\right)=4-8A\left(A+0,5\right)=4-8A^2-4A=-8A^2-4A+4\)

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

Để (1) có nghiệm thì Δ>=0

=>-4(A+1)(2A-1)>=0

=>(A+1)(2A-1)<=0

=>\(-1\le A\le\frac12\)

=>\(A\ge-1\)

=>\(\frac{2x-0,5}{2x^2+1}\ge-1\forall x\)

=>\(\frac{2x-0,5}{2x^2+1}+1,5\ge-1+1,5=0,5\forall x\)

Dấu '=' xảy ra khi \(\frac{2x-0,5}{2x^2+1}=-1\)

=>\(2x^2+1=-2x+0,5\)

=>\(2x^2+2x+0,5=0\)

=>\(4x^2+4x+1=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\)

\(B=3x^2-2x+7\\ =3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)+\dfrac{20}{3}\\ =3\left(x-\dfrac{1}{3}\right)^2+\dfrac{20}{3}\\ Vì:\left(x-\dfrac{1}{3}\right)^2\ge0\Rightarrow3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\in R\\ Vậy:min_B=\dfrac{20}{3}khi.\left(x-\dfrac{1}{3}\right)=0\Leftrightarrow x=\dfrac{1}{3}\)

vi 2x^2 lon hon hoac =0

 suy ra de p có gtnn thi 2x^2=0 

2x=0

x=0

suy ra p có gtnn là -5

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

Ta có: A = 25 - |3x - 6| - |3x + 8|

A = 25 - (|6 - 3x| + |3x + 8|) < = 25 - |6 - 3x + 3x + 8| = 25 - |14| = 25 - 14 = 11

Dấu "=" xảy ra <=> (3x - 6)(3x + 8) = 0

=> -8/3 \(\le\)x \(\le\)2

Vậy Max của A = 11 tại \(-\frac{8}{3}\le x\le2\)

Ta có: B = |2x - 5| - |2x - 11| + 3 > = |2x - 5 - 2x + 11| + 3 = |6| + 3  = 6 + 3 = 9

Dấu "=" xảy ra <=> (2x - 5)(2x - 11) = 0

=> \(\frac{5}{2}\le x\le\frac{11}{2}\)

Vậy Min của B = 9 tại \(\frac{5}{2}\le x\le\frac{11}{2}\)